JEE Mathematics Circle MCQs Set 05

Answer: (c) \( 2\sqrt2 \)

Explanation:
1. The points (1, 2) and (5, 2) lie on the line y = 2, and (5, 2), (5, −2) lie on x = 5. The angle at (5, 2) is a right angle.
2. So the hypotenuse joining (1, 2) and (5, −2) is a diameter: \( d=\sqrt{4^2+4^2}=4\sqrt2 \).
3. Radius \( =2\sqrt2 \).

Teacher's Note:
1. A right angle at a point of the circle means the opposite side is a diameter.
2. Check with the general form: the circle is x² + y² − 6x + 1 = 0, whose radius is √(9 − 1) = 2√2.

Answer: (a) x² + y² − 5x − 9y + 26 = 0

Explanation:
1. Centres: (3, 4) and (2, 5).
2. Circle on the diameter joining (x₁, y₁), (x₂, y₂): (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.
3. (x − 3)(x − 2) + (y − 4)(y − 5) = 0, i.e. x² + y² − 5x − 9y + 26 = 0.

Teacher's Note:
1. Use the diameter-form equation directly instead of finding the centre and radius separately.
2. Expand carefully: the constant term is 6 + 20 = 26.

Answer: (c) x² − ax + b = 0

Explanation:
1. Circle: (x − 0)(x − a) + (y − 1)(y − b) = 0.
2. Put y = 0: x(x − a) + (−1)(−b) = 0.
3. x² − ax + b = 0.

Teacher's Note:
1. Cutting the x-axis means substituting y = 0 in the circle equation.
2. The product of (y − 1)(y − b) at y = 0 is (−1)(−b) = +b.

Answer: (c) 9, 15

Explanation:
1. Here g = −5/2, f = −13/2, c = −14.
2. x-intercept \( =2\sqrt{g^2-c}=2\sqrt{\frac{25}{4}+14}=2\cdot\frac92=9 \).
3. y-intercept \( =2\sqrt{f^2-c}=2\sqrt{\frac{169}{4}+14}=2\cdot\frac{15}{2}=15 \).

Teacher's Note:
1. Intercept on x-axis is 2√(g² − c) and on y-axis is 2√(f² − c).
2. Be careful with the sign of c: here c is negative, so it adds to g² and f².

Answer: (c) 3x + 2y − 6 = 0

Explanation:
1. On the x-axis (y = 0): x² − 4x = 0, so the chord runs from (0, 0) to (4, 0); its mid-point is (2, 0).
2. On the y-axis (x = 0): y² − 6y = 0, so the chord runs from (0, 0) to (0, 6); its mid-point is (0, 3).
3. Line through (2, 0) and (0, 3): x/2 + y/3 = 1, i.e. 3x + 2y − 6 = 0.

Teacher's Note:
1. Find the end-points of each intercept first, then take mid-points.
2. Intercept form x/a + y/b = 1 gives the line at once.

Answer: (a) x² − y² = a² − b²

Explanation:
1. Let the centre be (h, k). The chord AB on the x-axis has length 2a, so the perpendicular distance from the centre to it is k and \( r^2=k^2+a^2 \).
2. Similarly the chord CD on the y-axis gives \( r^2=h^2+b^2 \).
3. Equating: k² + a² = h² + b², so h² − k² = a² − b².
4. Locus: x² − y² = a² − b².

Teacher's Note:
1. For a chord of length 2l at distance p from the centre, r² = p² + l².
2. Apply this to the two chords along the axes and equate the radii.

Answer: (c) A is true, but R is false.

Explanation:
1. Statement A: put y = 0, then x² − 2x = 0, so x = 0 or 2. The intercept length is 2, so A is true.
2. Statement R: the circle passes through the origin and its centre is (α/2, β/2), but its radius is \( \sqrt{\frac{\alpha^2}{4}+\frac{\beta^2}{4}}=\frac{\sqrt{\alpha^2+\beta^2}}{2} \), not \( \sqrt{\frac{\alpha^2+\beta^2}{2}} \). So R is false.
3. A is true and R is false.

Teacher's Note:
1. The radius is the distance from the centre to the origin, which is √(α²/4 + β²/4).
2. Always recompute the radius from the centre rather than trusting the stated form.

Answer: (c) (−6, −7)

Explanation:
1. Centre C = (−2, −3), radius \( \sqrt{4+9+5}=3\sqrt2 \).
2. Distance of C from the line: \( \frac{|-2-3+13|}{\sqrt2}=4\sqrt2 \), which is greater than the radius, so the line lies outside the circle.
3. The nearest point on the line to the circle is the foot of the perpendicular from C: move from C along the direction (−1, −1) by 4: (−2, −3) − 4(1, 1) = (−6, −7).

Teacher's Note:
1. The point of a line nearest to a circle outside it is the foot of the perpendicular from the centre.
2. The foot of the perpendicular from (x₁, y₁) to ax + by + c = 0 is (x₁, y₁) − (ax₁ + by₁ + c)/(a² + b²) · (a, b).

Answer: (b) 1

Explanation:
1. Substitute (8, 6): 64 + 36 − 100 = 0.
2. So the point lies on the circle.
3. Exactly one tangent can be drawn from a point on the circle.

Teacher's Note:
1. Compute S₁ first: positive gives 2 tangents, zero gives 1, negative gives none.
2. Here S₁ = 0, so the point is on the circle.

Answer: (d) y = 3, 12x + 5y = 39

Explanation:
1. A chord of length 8 in a circle of radius 5 is at distance \( \sqrt{25-16}=3 \) from the centre (origin).
2. Line through (2, 3) with slope m: mx − y + 3 − 2m = 0. Distance from the origin: \( \frac{|3-2m|}{\sqrt{1+m^2}}=3 \).
3. (3 − 2m)² = 9(1 + m²) gives 5m² + 12m = 0, so m = 0 or m = −12/5.
4. m = 0 gives y = 3. m = −12/5 gives 12x + 5y = 39. (The vertical line x = 2 gives chord 2√21, not 8.)

Teacher's Note:
1. Convert the chord length into the perpendicular distance from the centre using r² = d² + (l/2)².
2. Do not forget to test the vertical line separately, since its slope is undefined.

Answer: (d) a diameter

Explanation:
1. Centre of the circle is (2, −3).
2. Substitute into the line: 3(2) + 5(−3) + 9 = 6 − 15 + 9 = 0.
3. The line passes through the centre, so it is a diameter.

Teacher's Note:
1. Test whether the centre satisfies the line equation before doing any distance calculation.
2. A line through the centre always divides the circumference equally.

Answer: (c) 3

Explanation:
1. First circle: centre (1, 3), radius \( \sqrt{1+9-6}=2 \). A diameter is a chord of length 4, with its mid-point at (1, 3).
2. Distance between the centres: \( \sqrt{(2-1)^2+(1-3)^2}=\sqrt5 \). The line joining the centres is perpendicular to the chord at its mid-point, so by Pythagoras
3. \( R^2=(\sqrt5)^2+2^2=9 \).
4. Hence R = 3.

Teacher's Note:
1. The centre of the second circle lies on the perpendicular bisector of the chord, and the chord's mid-point is the centre of the first circle.
2. Use the right triangle: R² = (distance between centres)² + (half-chord)².

Answer: (b) 4x + 3y − 31 = 0, 4x + 3y + 19 = 0

Explanation:
1. Centre (3, −2), radius \( \sqrt{9+4+12}=5 \).
2. Tangent 4x + 3y + k = 0 must be at distance 5 from the centre: \( \frac{|12-6+k|}{5}=5 \).
3. |6 + k| = 25, so k = 19 or k = −31.
4. Tangents: 4x + 3y + 19 = 0 and 4x + 3y − 31 = 0.

Teacher's Note:
1. Parallel lines share the coefficients of x and y, so only the constant needs to be found.
2. Set perpendicular distance from the centre equal to the radius.

Answer: (c) g² + f² = c + k²

Explanation:
1. Centre (−g, −f). Distance from the centre to the line: the left side vanishes at the centre, so the distance is |0 − k| = |k| (the normal form has unit coefficients).
2. Radius \( =\sqrt{g^2+f^2-c} \).
3. Tangency: g² + f² − c = k², i.e. g² + f² = c + k².

Teacher's Note:
1. The line is already in normal form since cos²θ + sin²θ = 1, so no division is needed.
2. Tangent means distance from centre equals radius.

Answer: (b) (3, −1)

Explanation:
1. Tangent at (1, −2) to the first circle: x − 2y = 5.
2. Second circle: centre (4, −3), radius \( \sqrt{16+9-20}=\sqrt5 \). Distance of the centre from x − 2y − 5 = 0 is |4 + 6 − 5|/√5 = √5, so the line is tangent.
3. The point of contact is the foot of the perpendicular from (4, −3): (4, −3) − (5/5)(1, −2) = (3, −1).

Teacher's Note:
1. Check that the line really touches the second circle by comparing distance and radius.
2. The contact point is the foot of the perpendicular from the centre to the tangent.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. Centre (7, −1), radius 5. Distance from the origin to the centre: \( \sqrt{49+1}=5\sqrt2 \).
2. If 2α is the angle between the tangents, \( \sin\alpha=\frac{r}{d}=\frac{5}{5\sqrt2}=\frac1{\sqrt2} \), so α = π/4.
3. Angle between tangents = 2α = π/2.

Teacher's Note:
1. Draw the right triangle with the centre, the external point and a point of contact.
2. Also, S₁ = 49 + 1 − 25 = 25 equals r², so the tangent length equals the radius, confirming a right angle.

Answer: (c) (x + g)(x − 2) + (y + f)(y − 1) = 0

Explanation:
1. Angles APC and AQC (C the centre) are right angles, so A, P, Q, C lie on a circle with AC as diameter.
2. This is the circumcircle of triangle APQ.
3. Centre C = (−g, −f), A = (2, 1): (x − 2)(x + g) + (y − 1)(y + f) = 0.

Teacher's Note:
1. The radius to a point of contact is perpendicular to the tangent, so AC subtends a right angle at P and Q.
2. Use the diameter form with the end-points A and the centre.

Answer: (b) 12

Explanation:
1. The square of the tangent length equals the power of the point: PT² = PA · PB.
2. PT² = 16 × 9 = 144.
3. PT = 12.

Teacher's Note:
1. Power of a point: PA · PB = PT² for any secant through P.
2. Take the secant lengths measured from P, not the chord length AB.

Answer: (b) \( \sqrt{q-p} \)

Explanation:
1. For a point on the first circle, x² + y² + 2gx + 2fy = −p.
2. Tangent length squared to the second circle is S₁ = (x² + y² + 2gx + 2fy) + q = −p + q.
3. Length = √(q − p).

Teacher's Note:
1. Tangent length from (x₁, y₁) is √S₁, where S₁ is the circle expression evaluated at the point.
2. The two circles are concentric, so the value is the same for every point.

Answer: (d) 2x + y − 3 = 0

Explanation:
1. A diameter that bisects a chord is perpendicular to the chord.
2. The chord lies on x − 2y − 3 = 0 (slope 1/2), so the diameter has slope −2 and passes through the centre (2, −1).
3. y + 1 = −2(x − 2), i.e. 2x + y − 3 = 0.

Teacher's Note:
1. The line from the centre to the mid-point of a chord is perpendicular to the chord.
2. Use the centre and the negative reciprocal slope.

Answer: (a) \( x^2+y^2=\frac{4a^2}{3} \)

Explanation:
1. Tangents at P(α) and Q(β) meet at \( \left(\frac{a\cos\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}},\frac{a\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}}\right) \).
2. Its distance from the centre is \( \frac{a}{\cos\frac{\alpha-\beta}{2}}=\frac{a}{\cos(\pi/6)}=\frac{2a}{\sqrt3} \).
3. Hence x² + y² = 4a²/3.

Teacher's Note:
1. The meeting point of tangents at α and β lies on the line through the origin at angle (α+β)/2, at distance a/cos((α−β)/2).
2. Square the distance to get the locus.

Answer: (a) 3x + y − 4 = 0

Explanation:
1. Centre of the circle is (2, −2). Every normal passes through the centre.
2. Line through (1, 1) and (2, −2): slope = (−2 − 1)/(2 − 1) = −3.
3. y − 1 = −3(x − 1), i.e. 3x + y − 4 = 0.

Teacher's Note:
1. A normal to a circle is a line through its centre.
2. So the required normal is simply the line joining (1, 1) to the centre.