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

Question 1. In figure, PT is a tangent to a circle. If \( \angle BTA = 45^\circ \) and \( \angle PTB = 75^\circ \), find \( \angle ABT \).
Answer:

In the given figure, PQ is the tangent to the circle at point T.
We are given that \( \angle BTP = 75^\circ \) and \( \angle ATB = 45^\circ \).
The angles on a straight line add up to \( 180^\circ \). So, \( \angle QTA + \angle ATB + \angle BTP = 180^\circ \).
Substitute the known values:
\( \angle QTA + 45^\circ + 75^\circ = 180^\circ \)
\( \angle QTA + 120^\circ = 180^\circ \)
\( \implies \)
\( \angle QTA = 180^\circ - 120^\circ \)
\( \implies \)
\( \angle QTA = 60^\circ \)
Now, QTP is the tangent and TA is the chord of the circle. A chord creates an angle with the tangent at the point of contact.
The angle between the tangent and the chord (tangent-chord theorem) is equal to the angle in the alternate segment. This means \( \angle QTA = \angle ABT \).
\( \implies \)
\( \angle ABT = 60^\circ \)
In simple words: We find the missing angle on the straight line, then use a special rule that says the angle a tangent makes with a chord is the same as the angle in the opposite part of the circle.

🎯 Exam Tip: Remember the tangent-chord theorem which states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. This is a key property in circle geometry.

Question 2. In figure, TAS is a tangent to the circle, with centre O, at the point A. If \( \angle OBA = 32^\circ \), find the values of x and y.
Answer:

In the given figure, TAS is a tangent to the circle at point A, and O is the center.
We are given that \( \angle OBA = 32^\circ \).
Consider triangle OAB. OA and OB are both radii of the same circle, so \( OA = OB \).
This means triangle OAB is an isosceles triangle, so \( \angle OAB = \angle OBA \).
Therefore, \( \angle OAB = 32^\circ \).
The sum of angles in a triangle is \( 180^\circ \), so in \( \triangle AOB \):
\( \angle AOB + \angle OAB + \angle OBA = 180^\circ \)
\( \angle AOB + 32^\circ + 32^\circ = 180^\circ \)
\( \angle AOB + 64^\circ = 180^\circ \)
\( \implies \)
\( \angle AOB = 180^\circ - 64^\circ \)
\( \implies \)
\( \angle AOB = 116^\circ \)
Now, the arc AB makes an angle \( \angle AOB \) at the center and an angle \( \angle ACB \) at the remaining part of the circle. The angle at the center is twice the angle at the circumference.
So, \( \angle AOB = 2 \angle ACB \).
We are given that \( \angle ACB = y \).
\( 116^\circ = 2y \)
\( \implies \)
\( y = \frac{116^\circ}{2} \)
\( \implies \)
\( y = 58^\circ \)
Next, ST is a tangent and AB is a chord. According to the tangent-chord theorem, the angle between the tangent (ST) and the chord (AB) at the point of contact (A) is equal to the angle in the alternate segment. This means \( \angle BAS = \angle ACB \).
We are given that \( \angle BAS = x \).
So, \( x = \angle ACB \).
Since \( y = \angle ACB \), we have \( x = y \).
\( \implies \)
\( x = 58^\circ \)
Therefore, \( x = 58^\circ \) and \( y = 58^\circ \). Angles \( \angle OAB \) and \( \angle BAS \) together form \( \angle OAS \), which is \( 90^\circ \) because the radius is perpendicular to the tangent at the point of contact.
In simple words: First, we find the angles in the triangle formed by the center and two points on the circle, knowing that two sides are equal (radii). Then, we use the rule that an angle at the center is double the angle at the circumference. Finally, we use the rule that the angle between a tangent and a chord is the same as the angle in the alternate segment to find the value of x.

🎯 Exam Tip: Always remember that radii in the same circle are equal, creating isosceles triangles. Also, the angle a tangent makes with the radius at the point of contact is \( 90^\circ \).

Question 3. In figure, KLMN is a cyclic quadrilateral and PQ is a tangent to the circle at K. If LN is a diameter of the circle, \( \angle KLN = 30^\circ \) and \( \angle MNL = 60^\circ \), determine
(i) \( \angle QKN \)
(ii) \( \angle PKL \)
(iii) \( \angle MLK \)
Answer:

In the circle with center O, LN is the diameter.
We are given \( \angle KLN = 30^\circ \) and \( \angle MNL = 60^\circ \).
**Part (i) Determine \( \angle QKN \):**
Consider triangle LKN. Since LN is the diameter, the angle subtended by the diameter in a semicircle is a right angle.
So, \( \angle LKN = 90^\circ \).
In \( \triangle LKN \), the sum of angles is \( 180^\circ \).
\( \angle KLN + \angle LKN + \angle KNL = 180^\circ \)
\( 30^\circ + 90^\circ + \angle KNL = 180^\circ \)
\( 120^\circ + \angle KNL = 180^\circ \)
\( \implies \)
\( \angle KNL = 180^\circ - 120^\circ \)
\( \implies \)
\( \angle KNL = 60^\circ \)
Now, PQ is a tangent and KN is a chord at the point of contact K. Using the tangent-chord theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment.
So, \( \angle QKN = \angle KLN \).
Therefore, \( \angle QKN = 30^\circ \).
**Part (ii) Determine \( \angle PKL \):**
Again, PQ is a tangent and KL is a chord at the point of contact K.
Using the tangent-chord theorem, \( \angle PKL = \angle KNL \).
From part (i), we found \( \angle KNL = 60^\circ \).
Therefore, \( \angle PKL = 60^\circ \).
**Part (iii) Determine \( \angle MLK \):**
KLMN is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \).
So, \( \angle KNM + \angle KLM = 180^\circ \).
We can write \( \angle KNM \) as \( \angle KNL + \angle LNM \).
\( (\angle KNL + \angle LNM) + \angle MLK = 180^\circ \)
We know \( \angle KNL = 60^\circ \) and we are given \( \angle LNM = 60^\circ \).
\( 60^\circ + 60^\circ + \angle MLK = 180^\circ \)
\( 120^\circ + \angle MLK = 180^\circ \)
\( \implies \)
\( \angle MLK = 180^\circ - 120^\circ \)
\( \implies \)
\( \angle MLK = 60^\circ \)
In simple words: For (i), we use the angle in a semicircle rule and then the tangent-chord theorem. For (ii), we again use the tangent-chord theorem. For (iii), since it's a cyclic quadrilateral, we know that opposite angles add up to \( 180^\circ \). We find the full angle KNM and use it to calculate the last angle.

🎯 Exam Tip: Remember these key circle theorems: angle in a semicircle is \( 90^\circ \), opposite angles of a cyclic quadrilateral sum to \( 180^\circ \), and the tangent-chord theorem. Clearly state which theorem you are using for each step.

Question 4. In figure, AT is a tangent to the circle. If \( \angle ABC = 50^\circ \) and AC = BC, find \( \angle BAT \).
Answer:

In the given figure, AT is a tangent to the circle.
We are given \( \angle ABC = 50^\circ \) and \( AC = BC \).
Since \( AC = BC \) in \( \triangle ABC \), it is an isosceles triangle. This means the angles opposite to these equal sides are also equal.
So, \( \angle BAC = \angle ABC \).
Therefore, \( \angle BAC = 50^\circ \).
The sum of angles in a triangle is \( 180^\circ \). In \( \triangle ABC \):
\( \angle ACB + \angle BAC + \angle ABC = 180^\circ \)
\( \angle ACB + 50^\circ + 50^\circ = 180^\circ \)
\( \angle ACB + 100^\circ = 180^\circ \)
\( \implies \)
\( \angle ACB = 180^\circ - 100^\circ \)
\( \implies \)
\( \angle ACB = 80^\circ \)
Now, AT is the tangent and AB is the chord at the point of contact A. By the tangent-chord theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment.
So, \( \angle BAT = \angle ACB \).
Therefore, \( \angle BAT = 80^\circ \). The property of isosceles triangles simplifies the calculation of angles.

🎯 Exam Tip: Always look for isosceles triangles in circle geometry problems, as equal sides imply equal opposite angles. This often provides crucial information for solving the problem.

Question 5. In figure, O is the centre of the circumcircle of \( \triangle XYZ \). Tangents at X and Y intersects at T. Given, \( \angle XTY = 80^\circ \) and \( \angle XOZ = 140^\circ \). Calculate \( \angle ZXY \).
Answer:

In the figure, O is the center of the circumcircle of \( \triangle XYZ \).
Tangents XT and YT are drawn to the circle at points X and Y respectively, and they meet at T.
We are given \( \angle XTY = 80^\circ \) and \( \angle XOZ = 140^\circ \).
For tangents drawn from an external point to a circle, the angle between the tangents and the angle subtended by the chord joining the points of contact at the center are supplementary. So, \( \angle XTY + \angle XOY = 180^\circ \).
Substitute the value of \( \angle XTY \):
\( 80^\circ + \angle XOY = 180^\circ \)
\( \implies \)
\( \angle XOY = 180^\circ - 80^\circ \)
\( \implies \)
\( \angle XOY = 100^\circ \)
The angles around a point sum to \( 360^\circ \). So, around point O:
\( \angle XOY + \angle YOZ + \angle ZOX = 360^\circ \)
Substitute the known values:
\( 100^\circ + \angle YOZ + 140^\circ = 360^\circ \)
\( 240^\circ + \angle YOZ = 360^\circ \)
\( \implies \)
\( \angle YOZ = 360^\circ - 240^\circ \)
\( \implies \)
\( \angle YOZ = 120^\circ \)
Now, arc YZ subtends \( \angle YOZ \) at the center and \( \angle ZXY \) at the remaining part of the circle. The angle at the center is twice the angle at the circumference.
So, \( \angle YOZ = 2 \angle ZXY \).
\( 120^\circ = 2 \angle ZXY \)
\( \implies \)
\( \angle ZXY = \frac{120^\circ}{2} \)
\( \implies \)
\( \angle ZXY = 60^\circ \). This relationship between central and inscribed angles is fundamental in circle theorems.
In simple words: First, we find the angle at the center \( \angle XOY \) using the property of tangents from an external point. Then, we find \( \angle YOZ \) by subtracting the known angles around the center from \( 360^\circ \). Finally, we find \( \angle ZXY \) using the rule that the angle at the center is double the angle at the circumference.

🎯 Exam Tip: Remember the two key circle properties used here: (1) The angle formed by two tangents from an external point and the angle subtended by the chord of contact at the center are supplementary (sum to \( 180^\circ \)). (2) The angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle.

Question 6. In figure, O is the centre of the circle and AB is a chord of the circle. Line QBS is a tangent to the circle at B. If \( \angle AOB = 110^\circ \), find \( \angle APB \) and \( \angle ABQ \).
Answer:

O is the center of the circle, AB is a chord, and QBS is a tangent to the circle at point B.
We are given \( \angle AOB = 110^\circ \).
**Find \( \angle APB \):**
Arc AB subtends \( \angle AOB \) at the center and \( \angle APB \) at the remaining part of the circle (circumference). The angle at the center is twice the angle at the circumference.
So, \( \angle AOB = 2 \angle APB \).
\( 110^\circ = 2 \angle APB \)
\( \implies \)
\( \angle APB = \frac{110^\circ}{2} \)
\( \implies \)
\( \angle APB = 55^\circ \)
**Find \( \angle ABQ \):**
QBS is the tangent and AB is the chord at the point of contact B. By the tangent-chord theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment.
So, \( \angle ABQ = \angle APB \).
Since \( \angle APB = 55^\circ \), then \( \angle ABQ = 55^\circ \). These theorems are essential for solving problems involving tangents and chords.
In simple words: First, we use the rule that the angle at the center of a circle is double the angle at the circumference to find \( \angle APB \). Then, we use the tangent-chord theorem, which states that the angle between a tangent and a chord is equal to the angle in the opposite segment of the circle, to find \( \angle ABQ \).

🎯 Exam Tip: When dealing with tangents and chords, always consider both the relationship between central and inscribed angles, and the tangent-chord theorem. These two theorems are frequently used together in such problems.

Question 7. In figure, AB is a diameter and AC is a chord of a circle such that \( \angle BAC = 30^\circ \). The tangent at C intersects AB produced at D. Prove that BC = BD.
Answer:

Given: A circle with center O, AB is its diameter, and C is a point on the circle. A tangent is drawn at C, meeting AB produced at D. We are given \( \angle BAC = 30^\circ \).
To prove: \( BC = BD \).
Construction: Join BC.
Proof:
Since CD is a tangent and CB is a chord, by the tangent-chord theorem, the angle between the tangent (CD) and the chord (CB) is equal to the angle in the alternate segment.
So, \( \angle DCB = \angle BAC \).
Since \( \angle BAC = 30^\circ \), we have \( \angle DCB = 30^\circ \).
Now consider \( \triangle ABC \). Since AB is the diameter, the angle subtended by the diameter in a semicircle is \( 90^\circ \).
So, \( \angle ACB = 90^\circ \).
The sum of angles in \( \triangle ABC \) is \( 180^\circ \).
\( \angle BAC + \angle CBA + \angle ACB = 180^\circ \)
\( 30^\circ + \angle CBA + 90^\circ = 180^\circ \)
\( 120^\circ + \angle CBA = 180^\circ \)
\( \implies \)
\( \angle CBA = 180^\circ - 120^\circ \)
\( \implies \)
\( \angle CBA = 60^\circ \)
Now, consider \( \triangle BCD \). The exterior angle \( \angle CBA \) of \( \triangle BCD \) is equal to the sum of the interior opposite angles.
So, \( \angle CBA = \angle BCD + \angle BDC \).
\( 60^\circ = 30^\circ + \angle BDC \)
\( \implies \)
\( \angle BDC = 60^\circ - 30^\circ \)
\( \implies \)
\( \angle BDC = 30^\circ \)
In \( \triangle BCD \), we have \( \angle BCD = 30^\circ \) and \( \angle BDC = 30^\circ \).
Since two angles are equal, the sides opposite to these angles must also be equal.
Therefore, \( BC = BD \) (Sides opposite to equal angles).
Hence proved. Understanding the exterior angle theorem for triangles is key here.
In simple words: First, we use the tangent-chord theorem to find \( \angle DCB \). Then, we use the angle in a semicircle rule to find \( \angle ACB \), and then find \( \angle CBA \) from \( \triangle ABC \). Finally, we use the exterior angle property of \( \triangle BCD \) to find \( \angle BDC \). Since \( \angle BCD \) and \( \angle BDC \) are equal, the sides opposite to them, BC and BD, must also be equal.

🎯 Exam Tip: For proof-based questions, clearly state your given information, what you need to prove, and any constructions made. Use precise geometric reasons for each step, such as "angle in a semicircle," "tangent-chord theorem," or "exterior angle property of a triangle."

Question 8. In figure, DE is a tangent to the circumcircle of \( \triangle ABC \) at the vertex A such that DE || BC. Show that AB = AC.
Answer:

Given: A circle with a circumcircle of \( \triangle ABC \). DE is a tangent to the circle at vertex A, and \( DE \parallel BC \).
To prove: \( AB = AC \).
Proof:
Since DE is the tangent and AB is a chord of the circle at the point of contact A, by the tangent-chord theorem:
\( \angle DAB = \angle ACB \) (Angle in the alternate segment) — (i)
We are given that \( DE \parallel BC \). Since AB is a transversal intersecting parallel lines DE and BC, the alternate interior angles are equal.
So, \( \angle DAB = \angle ABC \) (Alternate angles) — (ii)
From equations (i) and (ii), we have:
\( \angle ACB = \angle ABC \)
In \( \triangle ABC \), since \( \angle ACB = \angle ABC \), the sides opposite to these equal angles must also be equal. When two angles in a triangle are equal, the triangle is isosceles.
Therefore, \( AB = AC \) (Sides opposite equal angles).
Hence proved. This demonstrates how parallel lines and tangents relate within a circle.
In simple words: First, we use the tangent-chord theorem to show that \( \angle DAB \) is equal to \( \angle ACB \). Then, because DE is parallel to BC, we know that \( \angle DAB \) is also equal to \( \angle ABC \) (alternate angles). Since both \( \angle ACB \) and \( \angle ABC \) are equal to \( \angle DAB \), they must be equal to each other. If two angles in a triangle are equal, the sides opposite those angles are also equal, proving that \( AB = AC \).

🎯 Exam Tip: When parallel lines are involved with circles and tangents, always look for alternate interior angles or corresponding angles to establish relationships between angles. Combine this with circle theorems to reach your proof.

Question 9. In figure, two circles intersect in B and C. Lines ABD and ACE are drawn to meet the second circle at points D and E, AF is a tangent at A. Prove that AF || DE. [Hint Join B to C]
Answer:

Given: Two circles intersect each other at points B and C. Lines ABD and ACE are drawn to meet the second circle at points D and E respectively. AF is a tangent to the first circle at A.
To prove: \( AF \parallel DE \).
Construction: Join BC.
Proof:
In the first circle, AF is the tangent at A and AB is the chord. By the tangent-chord theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment.
So, \( \angle BAF = \angle ACB \) — (i)
Now consider the cyclic quadrilateral BCED in the second circle. For a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
So, \( \angle ACB = \angle BDE \) — (ii)
From equations (i) and (ii), we have:
\( \angle BAF = \angle BDE \)
These two angles, \( \angle BAF \) and \( \angle BDE \), are alternate angles formed by the transversal AD intersecting lines AF and DE.
Since the alternate angles are equal, the lines AF and DE must be parallel.
Therefore, \( AF \parallel DE \). This problem effectively links tangent-chord theorem with properties of cyclic quadrilaterals.
In simple words: First, we use the tangent-chord theorem in the first circle to say that \( \angle BAF \) is the same as \( \angle ACB \). Then, in the second circle, we see that BCED is a cyclic quadrilateral, so its exterior angle \( \angle ACB \) is equal to the interior opposite angle \( \angle BDE \). Since both \( \angle BAF \) and \( \angle BDE \) are equal to \( \angle ACB \), they must be equal to each other. Because they are alternate angles, lines AF and DE must be parallel.

🎯 Exam Tip: When working with two intersecting circles, drawing the common chord (like BC here) is often a crucial step. It helps connect properties between the two circles. Also, correctly identifying cyclic quadrilaterals and applying their properties is vital.

Question 10. In the figure, CD is the tangent line at C to the circumcircle of \( \triangle ABC \) intersecting AB produced in D. Show that \( \triangle DBC \sim \triangle DCA \).
Answer:

Given: CD is the tangent line at C to the circumcircle of \( \triangle ABC \). AB is produced to meet the tangent at D.
To show: \( \triangle DBC \sim \triangle DCA \).
Proof:
Since CD is the tangent to the circle at C and BC is a chord, by the tangent-chord theorem, the angle between the tangent (CD) and the chord (BC) is equal to the angle in the alternate segment.
So, \( \angle BCD = \angle BAC \). (This is also written as \( \angle DAC \))
Now consider \( \triangle DBC \) and \( \triangle DCA \).
1. \( \angle D = \angle D \) (Common angle to both triangles).
2. \( \angle BCD = \angle BAC \) (Proved above, using the tangent-chord theorem).
Since two angles of \( \triangle DBC \) are equal to two corresponding angles of \( \triangle DCA \), the triangles are similar by the Angle-Angle (AA) similarity criterion.
Therefore, \( \triangle DBC \sim \triangle DCA \). This proves similarity through angle relationships.
In simple words: We want to show that two triangles, DBC and DCA, are similar. First, we notice that both triangles share the angle at D. Next, we use a rule about tangents and chords in a circle: the angle formed by the tangent CD and the chord BC (which is \( \angle BCD \)) is equal to the angle in the alternate segment, \( \angle BAC \). Since we have two pairs of equal angles, the triangles are similar.

🎯 Exam Tip: For similarity proofs involving circles and tangents, the tangent-chord theorem is often key to establishing an angle relationship. Always look for common angles between the triangles, as this provides one of the necessary conditions for AA similarity.