OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Exercise 14 (D)

 

Question 1. In figure, PA and PB are tangents from P to a circle with centre O. At M, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN.
Answer:

Answer:Given: We have a circle with center O. From an outside point P, PA and PB are tangents to the circle. A new tangent is drawn at point M on the circle, intersecting PA at K and PB at N. To prove: \( KN = AK + BN \) Proof: From point K, KA and KM are both tangents drawn to the circle from K. Therefore, \( KA = KM \) ...(i) This is because tangents drawn from an external point to a circle are equal in length. Similarly, from point N, NM and NB are tangents drawn to the circle from N. Therefore, \( NB = MN \) ...(ii) Now, we add equations (i) and (ii): \( KA + NB = KM + MN \) We can see from the figure that \( KM + MN \) forms the line segment \( KN \). So, \( KN = KA + NB \). Hence proved.In simple words: When you draw lines from a point outside a circle that just touch the circle (these are called tangents), the lengths of these lines from that point to the circle are always the same. We use this rule for point K and point N to show that the length KN is equal to the sum of AK and BN.

🎯 Exam Tip: The key to solving tangent problems is remembering that tangents drawn from an external point to a circle are always equal in length. Clearly state this theorem as a reason for each step in your proof.

 

Question 2. In figure, two concentric circles are given. Prove that every chord of the greater circle which is tangent to the smaller circle is bisected at its point of contact.
Answer:

Answer:Given: We have two circles with the same center O. A chord AB of the larger circle touches the smaller circle at point P. To prove: AB is bisected at P, which means \( AP = PB \). Construction: We join OP, OA, and OB. Proof: OP is the radius of the smaller circle, and AB is a tangent to the smaller circle at P. We know that the radius drawn to the point of tangency is perpendicular to the tangent. So, \( OP \perp AB \). This means \( \angle OPA = 90^\circ \) and \( \angle OPB = 90^\circ \). Now, let's consider the two right-angled triangles, \( \triangle OAP \) and \( \triangle OBP \). 1. Hypotenuse OA = Hypotenuse OB (These are radii of the same larger circle). 2. Side OP = Side OP (This side is common to both triangles). So, by the RHS (Right angle-Hypotenuse-Side) congruence axiom, \( \triangle OAP \cong \triangle OBP \). Since the triangles are congruent, their corresponding parts are equal. Therefore, \( AP = PB \) (by C.P.C.T. - Corresponding Parts of Congruent Triangles). This means that P bisects the chord AB. Hence proved.In simple words: Imagine two circles, one inside the other, sharing the same middle point. If you draw a straight line across the big circle so it just touches the small circle, that line will be cut exactly in half at the point where it touches the small circle. This happens because the line from the center to where it touches the small circle is perpendicular, helping us form two identical triangles.

🎯 Exam Tip: When proving a chord is bisected by a tangent, always use the fact that the radius to the point of contact is perpendicular to the tangent. This often leads to using RHS congruence for right-angled triangles.

 

Question 3. Prove that the angle between two tangents drawn from an external point to a circle is supplementary to angle subtended by the line segments joining the points of contact at the centre.
Answer:

Answer:Given: We have a circle with center O. From an external point P, two tangents PT and PS are drawn. The line segment TS connects the points of contact, and it forms an angle \( \angle TOS \) at the center. To prove: \( \angle TPS + \angle TOS = 180^\circ \) (which means they are supplementary). Construction: Join OP. Proof: We know that the radius drawn to the point of tangency is perpendicular to the tangent. So, \( OT \perp PT \) and \( OS \perp PS \). This means \( \angle OTP = 90^\circ \) and \( \angle OSP = 90^\circ \). Now, consider the quadrilateral PTOS. The sum of all angles in a quadrilateral is \( 360^\circ \). So, \( \angle OTP + \angle TOS + \angle OSP + \angle TPS = 360^\circ \). Substitute the known right angles: \( 90^\circ + \angle TOS + 90^\circ + \angle TPS = 360^\circ \) \( 180^\circ + \angle TOS + \angle TPS = 360^\circ \) Now, subtract \( 180^\circ \) from both sides: \( \angle TOS + \angle TPS = 360^\circ - 180^\circ \) \( \implies \angle TPS + \angle TOS = 180^\circ \) Thus, \( \angle TPS \) and \( \angle TOS \) are supplementary. Hence proved.In simple words: When you draw two lines from a point outside a circle, and these lines just touch the circle, they are called tangents. The angle made by these two tangents, and the angle made by the lines from the center to where they touch, will always add up to 180 degrees. This is because these four points form a shape with four sides, and we know two of its angles are 90 degrees.

🎯 Exam Tip: Remember that a radius is always perpendicular to a tangent at the point of contact. This creates 90-degree angles, which is crucial when dealing with quadrilaterals involving tangents and radii from the center.

 

Question 4. In figure, a circle touches the side BC of a △ABC at X and AB and AC produced at Y and Z respectively. Prove that AY is half the perimeter of △ABC.
Answer:

Answer:Given: We have a triangle \( \triangle ABC \). A circle touches the side BC at point X. The sides AB and AC are extended (produced) to touch the circle at points Y and Z respectively. To prove: \( AY = \frac{1}{2} (AB + BC + CA) \) (which is half the perimeter of \( \triangle ABC \)). Proof: We know that tangents drawn from an external point to a circle are equal in length. 1. From point A, AY and AZ are tangents to the circle. So, \( AY = AZ \) ...(i) 2. From point B, BY and BX are tangents to the circle. So, \( BY = BX \) ...(ii) 3. From point C, CX and CZ are tangents to the circle. So, \( CZ = CX \) ...(iii) Now, let's consider the length AY. \( AY = AB + BY \) From (ii), we can substitute \( BY = BX \): \( AY = AB + BX \) ...(iv) Similarly, let's consider the length AZ. \( AZ = AC + CZ \) From (iii), we can substitute \( CZ = CX \): \( AZ = AC + CX \) ...(v) Now, let's add equations (iv) and (v): \( AY + AZ = (AB + BX) + (AC + CX) \) Rearrange the terms: \( AY + AZ = AB + AC + (BX + CX) \) From the figure, we can see that \( BX + CX \) makes up the side BC. So, \( AY + AZ = AB + AC + BC \) From (i), we know that \( AY = AZ \). So, we can replace AZ with AY: \( AY + AY = AB + AC + BC \) \( \implies 2AY = AB + BC + CA \) Now, divide both sides by 2: \( \implies AY = \frac{1}{2} (AB + BC + CA) \) This means AY is equal to half the perimeter of \( \triangle ABC \). This property is important for calculating the semi-perimeter of a triangle in some cases. Hence proved.In simple words: When a circle touches one side of a triangle and the extensions of the other two sides, the distance from the top corner (A) to where it touches the extended side (Y) is exactly half the total distance around the triangle. This is because the pieces of lines that touch the circle from the same outside point are always the same length.

🎯 Exam Tip: For problems involving excircles and tangents, always identify all pairs of equal tangents from external vertices. This is the fundamental property that allows you to express side lengths in terms of tangents, which can then be manipulated algebraically.

 

Question 5. In figure, the incircle of a △ABC touches the sides BC, CA and AB at D, E and F respectively. Show that AF + BD + CE = AE + CD + BF = \( \frac{1}{2} \) (perimeter of △ABC).
Answer:

Answer:Given: An incircle of \( \triangle ABC \) touches the sides BC, CA, and AB at points D, E, and F respectively. To prove: \( AF + BD + CE = AE + CD + BF = \frac{1}{2} (AB + BC + CA) \) Proof: We know that tangents drawn from an external point to a circle are equal in length. 1. From vertex A, AF and AE are tangents to the circle. So, \( AF = AE \) ...(i) 2. From vertex B, BD and BF are tangents to the circle. So, \( BD = BF \) ...(ii) 3. From vertex C, CE and CD are tangents to the circle. So, \( CE = CD \) ...(iii) Now, let's add equations (i), (ii), and (iii): \( AF + BD + CE = AE + BF + CD \) This proves the first part of the statement. For the second part, let's consider the perimeter of \( \triangle ABC \). Perimeter \( = AB + BC + CA \) We can write the sides in terms of the tangent segments: \( AB = AF + FB \) \( BC = BD + DC \) \( CA = CE + EA \) So, Perimeter \( = (AF + FB) + (BD + DC) + (CE + EA) \) Rearrange the terms: Perimeter \( = AF + BD + CE + FB + DC + EA \) From our previous result, we know \( AF + BD + CE = AE + BF + CD \). So, we can substitute \( (FB + DC + EA) \) with \( (BD + CE + AF) \) from the left side of the previous equation. Perimeter \( = (AF + BD + CE) + (AF + BD + CE) \) Perimeter \( = 2 (AF + BD + CE) \) Divide by 2: \( \implies AF + BD + CE = \frac{1}{2} (AB + BC + CA) \) Since \( AF + BD + CE = AE + CD + BF \), it also means: \( AE + CD + BF = \frac{1}{2} (AB + BC + CA) \) This shows that the sum of alternate tangent segments from the vertices is equal to half the perimeter of the triangle. Hence proved.In simple words: When a circle is drawn inside a triangle, just touching all three sides, it creates six small lines from the corners to the touch points. This proof shows that if you add up the lengths of three special lines (AF, BD, CE), it's the same as adding up the other three (AE, CD, BF). And both these sums are exactly half the total distance around the triangle.

🎯 Exam Tip: Always remember the fundamental property that tangents from an external point to a circle are equal. This allows you to set up equalities for tangent segments, which are crucial for proving relationships involving the perimeter of a triangle with an incircle.

 

Question 6. Two circles touch each other externally. Show that the lengths of the tangents, drawn to the two circles from any point on the common tangents, are equal (See figure).
Answer:

Answer:Given: Two circles touch each other externally at point T. A common tangent passes through T. From a point P on this common tangent, other tangents PA and PB are drawn to the two circles. To prove: \( PA = PB \) Proof: We know that tangents drawn from an external point to a circle are equal in length. 1. For the first circle (let's say the one on the left), P is an external point. PT and PA are tangents drawn from P to this circle. So, \( PA = PT \) ...(i) 2. For the second circle (the one on the right), P is also an external point. PT and PB are tangents drawn from P to this circle. So, \( PT = PB \) ...(ii) Now, by combining equations (i) and (ii): Since \( PA = PT \) and \( PT = PB \), it directly follows that \( PA = PB \). This demonstrates a useful property: if a point lies on the common tangent of two externally touching circles, then the tangents from that point to each circle are equal. Hence proved.In simple words: When two circles touch each other on the outside, and you have a point on the line that touches both circles at that meeting point, then any other lines you draw from that point to touch each circle will be the same length. It's like having a special point from which all touching lines have equal reach.

🎯 Exam Tip: This proof relies solely on the property that two tangents drawn from an external point to a circle are equal in length. Ensure you clearly identify which tangents belong to which circle and from which external point.

 

Question 7. Two circles touch externally at A. A common tangent touches them at B and C, and another common tangent at A meets the previous common tangent at P, as is shown in figure. Prove that
(i) PB = PC,
(ii) ∠BAC = 90°
Answer:

Answer:Given: Two circles touch each other externally at A. A common tangent touches them at B and C. Another common tangent, passing through A, meets the tangent BC at P. To prove: (i) \( PB = PC \) (ii) \( \angle BAC = 90^\circ \) Proof: (i) To prove \( PB = PC \): From point P, tangents PB and PA are drawn to the first circle (the one on the left). Therefore, \( PB = PA \) ...(1) Similarly, from point P, tangents PC and PA are drawn to the second circle (the one on the right). Therefore, \( PC = PA \) ...(2) From equations (1) and (2), since both PB and PC are equal to PA, we can conclude: \( PB = PC \) This means that the point P is equidistant from B, C, and A. (ii) To prove \( \angle BAC = 90^\circ \): Consider \( \triangle PBA \). Since \( PB = PA \) (from part i), \( \triangle PBA \) is an isosceles triangle. Angles opposite to equal sides in a triangle are equal. So, \( \angle PBA = \angle PAB \) ...(3) Similarly, consider \( \triangle PCA \). Since \( PC = PA \) (from part i), \( \triangle PCA \) is an isosceles triangle. So, \( \angle PCA = \angle PAC \) ...(4) Now, let's look at \( \triangle ABC \). The sum of angles in any triangle is \( 180^\circ \). \( \angle BAC + \angle ABC + \angle ACB = 180^\circ \) We know that \( \angle ABC = \angle PBA \) and \( \angle ACB = \angle PCA \). Substitute from (3) and (4): \( \angle BAC + \angle PAB + \angle PAC = 180^\circ \) Notice that \( \angle PAB + \angle PAC \) is the same as \( \angle BAC \). So, \( \angle BAC + \angle BAC = 180^\circ \) \( \implies 2 \angle BAC = 180^\circ \) Divide by 2: \( \implies \angle BAC = \frac{180^\circ}{2} \) \( \implies \angle BAC = 90^\circ \) Hence proved. The point of contact of two externally touching circles forms a right angle with the two points of tangency of a direct common tangent, when connected to the point of intersection of the tangents.In simple words: When two circles touch each other from the outside, and two lines just touch both circles, if those lines meet at a point P, then the distances from P to where they touch the circles are equal. Also, the angle formed by connecting the point where the circles touch each other (A) to the points where the other lines touch the circles (B and C) will always be a perfect right angle (90 degrees).

🎯 Exam Tip: This problem combines two critical tangent properties: equal tangents from an external point, and angle sum in a triangle. Clearly mark equal angles in your diagram to easily see the isosceles triangles, which simplifies the angle proof.

 

Question 8. Two circles touch internally at A. P is any point on the tangent at A. From P, two tangents PB and PC are drawn to the two circles. Prove that PB = PC.
Answer:

Answer:Given: Two circles touch internally at point A. P is a point on the common tangent drawn at A. From P, two tangents PB and PC are drawn to the two circles. To prove: \( PB = PC \) Proof: We know that tangents drawn from an external point to a circle are equal in length. 1. Consider the smaller circle. From point P, tangents PA and PB are drawn to this circle. So, \( PA = PB \) ...(i) 2. Consider the larger circle. From point P, tangents PA and PC are drawn to this circle. So, \( PA = PC \) ...(ii) From equations (i) and (ii): Since both PB and PC are equal to PA, we can conclude that \( PB = PC \). This demonstrates that if a point lies on the common tangent of two internally touching circles, then the lengths of the tangents from that point to each circle are equal. Hence proved.In simple words: Imagine a small circle inside a big circle, touching at one spot. If you draw a line that just touches both circles at that spot, and then pick any point (P) on that line, any other lines you draw from P to touch each circle will be the same length. It’s like P has the same "reach" to both circles.

🎯 Exam Tip: This proof is straightforward if you recall the fundamental theorem about tangents from an external point being equal. Clearly identify the external point (P) and the pairs of tangents for each circle (PA and PB for the inner; PA and PC for the outer).

 

Question 9. From a point P, outside a circle, with centre O, tangents PA and PB are drawn as shown in the figure. Prove that :
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of AB.
Answer:

Answer:Given: From an external point P, tangents PA and PB are drawn to a circle with center O. Lines OP, OA, OB, and AB are joined. To prove: (i) \( \angle AOP = \angle BOP \) (ii) OP is the perpendicular bisector of AB. Proof: (i) To prove \( \angle AOP = \angle BOP \): Consider the two right-angled triangles, \( \triangle OAP \) and \( \triangle OBP \). We know that the radius to the point of tangency is perpendicular to the tangent. So, \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \). 1. Hypotenuse OP = Hypotenuse OP (This side is common to both triangles). 2. Side OA = Side OB (These are radii of the same circle). So, by the RHS (Right angle-Hypotenuse-Side) congruence axiom, \( \triangle OAP \cong \triangle OBP \). Since the triangles are congruent, their corresponding parts are equal. Therefore, \( \angle AOP = \angle BOP \) (by C.P.C.T. - Corresponding Parts of Congruent Triangles). Also, \( \angle APO = \angle BPO \) (by C.P.C.T.). This means OP bisects \( \angle APB \). (ii) To prove OP is the perpendicular bisector of AB: Let M be the point where OP intersects AB. Now, consider \( \triangle OAM \) and \( \triangle OBM \). 1. OA = OB (Radii of the same circle). 2. OM = OM (Common side). 3. \( \angle AOM = \angle BOM \) (This is the same as \( \angle AOP = \angle BOP \) which we proved in part i). So, by the SAS (Side-Angle-Side) congruence criterion, \( \triangle OAM \cong \triangle OBM \). Since the triangles are congruent, their corresponding parts are equal. Therefore, \( AM = BM \) (by C.P.C.T.). This means M is the midpoint of AB, so OP bisects AB. Also, \( \angle OMA = \angle OMB \) (by C.P.C.T.). Since \( \angle OMA \) and \( \angle OMB \) form a linear pair (angles on a straight line), their sum is \( 180^\circ \). \( \angle OMA + \angle OMB = 180^\circ \) Since they are equal, \( 2 \angle OMA = 180^\circ \) \( \implies \angle OMA = 90^\circ \) Since OP passes through M and \( \angle OMA = 90^\circ \), OP is perpendicular to AB. Combining \( AM = BM \) (bisector) and \( OP \perp AB \) (perpendicular), we conclude that OP is the perpendicular bisector of AB. Hence proved. This establishes important symmetrical properties of tangents from an external point.In simple words: When you draw two lines from a point outside a circle that just touch the circle, and you also draw a line from the center (O) to that outside point (P), then: 1. The angle at the center formed by the two touch points is split equally by the line OP. 2. The line OP cuts the line connecting the two touch points (AB) exactly in half and at a perfect right angle.

🎯 Exam Tip: Always start by proving the congruence of \( \triangle OAP \) and \( \triangle OBP \) using RHS. This establishes the equality of angles at the center and at P, which are then used in the second set of congruent triangles to prove the perpendicular bisector property.

 

Question 10. Prove that the tangents to a circle at the extremities of any chord make equal angles with the chord.
Answer:

Answer:Given: In a circle with center O, AB is a chord. Tangents PA and PB are drawn at the extremities (ends) of the chord A and B, which meet at point P. To prove: \( \angle PAB = \angle PBA \) (The tangents make equal angles with the chord). Construction: Join OA, OB, and OP. Proof: Consider \( \triangle OAP \) and \( \triangle OBP \). 1. OP = OP (Common side). 2. OA = OB (Radii of the same circle). 3. PA = PB (Tangents drawn from an external point P to a circle are equal in length). So, by the SSS (Side-Side-Side) congruence criterion, \( \triangle OAP \cong \triangle OBP \). Since the triangles are congruent, their corresponding parts are equal. Therefore, \( \angle APO = \angle BPO \) (by C.P.C.T.). This means OP bisects \( \angle APB \). Now, let M be the point where OP intersects the chord AB. Consider \( \triangle PAM \) and \( \triangle PBM \). 1. PA = PB (Tangents from external point P are equal). 2. PM = PM (Common side). 3. \( \angle APM = \angle BPM \) (This is the same as \( \angle APO = \angle BPO \), which we just proved). So, by the SAS (Side-Angle-Side) congruence criterion, \( \triangle PAM \cong \triangle PBM \). Since the triangles are congruent, their corresponding parts are equal. Therefore, \( \angle PAM = \angle PBM \) (by C.P.C.T.). The angle \( \angle PAM \) is the angle between tangent PA and chord AB. The angle \( \angle PBM \) is the angle between tangent PB and chord AB. Thus, the tangents to a circle at the extremities of any chord make equal angles with the chord. Hence proved. This property shows the symmetry in how tangents interact with chords.In simple words: If you draw a line inside a circle (called a chord), and then draw two lines that just touch the circle at each end of the chord, those touching lines will make the same angle with the chord. For example, if the first touching line makes a 30-degree angle with the chord, the second touching line will also make a 30-degree angle with the chord.

🎯 Exam Tip: This proof involves two sets of triangle congruency. First, use SSS to show \( \triangle OAP \cong \triangle OBP \), leading to \( \angle APO = \angle BPO \). Then, use SAS with \( \triangle PAM \) and \( \triangle PBM \) to prove the main statement, being careful to use the correct vertices for the angles with the chord.

 

Question 11. In the figure, two circles touch each other externally at C. Prove that the common tangent at C bisects the other two common tangents.
Answer:

Answer:Given: Two circles with centers O and O' touch each other externally at C. PQ and RS are two direct common tangents. A third common tangent passes through C and intersects PQ at L and RS at M. To prove: LM bisects PQ and RS. This means L is the midpoint of PQ and M is the midpoint of RS. Proof: We know that tangents drawn from an external point to a circle are equal in length. 1. Consider point L on the tangent PQ. From L, LP and LC are tangents to the first circle (left one). So, \( LP = LC \) ...(i) 2. Also from point L, LQ and LC are tangents to the second circle (right one). So, \( LC = LQ \) ...(ii) From (i) and (ii), since both LP and LQ are equal to LC, we get: \( LP = LC = LQ \) This means \( LP = LQ \), so L is the midpoint of the tangent segment PQ. 3. Similarly, consider point M on the tangent RS. From M, MR and MC are tangents to the first circle. So, \( MR = MC \) ...(iii) 4. Also from point M, MS and MC are tangents to the second circle. So, \( MC = MS \) ...(iv) From (iii) and (iv), since both MR and MS are equal to MC, we get: \( MR = MC = MS \) This means \( MR = MS \), so M is the midpoint of the tangent segment RS. Therefore, the common tangent at C (the line segment LM) bisects both common tangents PQ and RS. This shows how tangent properties create symmetrical relationships. Hence proved.In simple words: Imagine two circles touching each other. If you draw a line that just touches both circles on the top, and another line that just touches them on the bottom, and then draw a third line that touches the circles exactly where they meet, this third line will cut the top line in half and also cut the bottom line in half. This is because any lines from an outside point that touch a circle have the same length.

🎯 Exam Tip: The core concept here is that tangents from an external point to a circle are equal. Apply this principle systematically for point L (with tangents LP, LC, LQ) and point M (with tangents MR, MC, MS) to show that L and M are midpoints.

 

Question 12. PA and PB are tangents drawn from an external point P to a circle with centre C (See figure). Prove that ∠APB = 2∠CAB.
Answer:

Answer:Given: PA and PB are two tangents drawn from an external point P to a circle with center C. CA is joined. To prove: \( \angle APB = 2 \angle CAB \) Proof: Consider \( \triangle APB \). We know that tangents drawn from an external point to a circle are equal in length. So, \( AP = BP \). Since \( AP = BP \), \( \triangle APB \) is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. So, \( \angle PAB = \angle PBA \) ...(i) The sum of angles in \( \triangle APB \) is \( 180^\circ \). \( \angle PAB + \angle PBA + \angle APB = 180^\circ \) Substitute \( \angle PBA \) with \( \angle PAB \) from (i): \( \angle PAB + \angle PAB + \angle APB = 180^\circ \) \( \implies 2 \angle PAB + \angle APB = 180^\circ \) Rearrange to find \( \angle APB \): \( \implies \angle APB = 180^\circ - 2 \angle PAB \) ...(ii) Now, let's look at \( \angle PAB \). We know that the radius to the point of tangency is perpendicular to the tangent. So, \( CA \perp PA \), which means \( \angle CAP = 90^\circ \). We can express \( \angle PAB \) as: \( \angle PAB = \angle CAP - \angle CAB \) Since \( \angle CAP = 90^\circ \): \( \angle PAB = 90^\circ - \angle CAB \) ...(iii) Now, substitute the expression for \( \angle PAB \) from (iii) into equation (ii): \( \angle APB = 180^\circ - 2 (90^\circ - \angle CAB) \) Distribute the -2: \( \implies \angle APB = 180^\circ - (2 \times 90^\circ) + (2 \times \angle CAB) \) \( \implies \angle APB = 180^\circ - 180^\circ + 2 \angle CAB \) \( \implies \angle APB = 2 \angle CAB \) Hence proved. This establishes a relationship between the angle formed by the tangents and the angle formed by the radius and chord.In simple words: When you draw two lines from a point (P) outside a circle that just touch it, the angle created by those two touching lines (APB) is twice as big as the angle created by the line from the center (C) to one touch point (A) and the line connecting the touch point to the point (B) on the other tangent (CAB). This is because the triangle formed by the two tangents and the chord is an isosceles triangle.

🎯 Exam Tip: This proof cleverly uses the isosceles triangle formed by the tangents (AP=BP) and the fact that the radius is perpendicular to the tangent. The key step is expressing \( \angle PAB \) as \( 90^\circ - \angle CAB \). Be careful with the angle names and subtractions.