JEE Mathematics Circles MCQs Set 19

Answer: (c) 5

Explanation:
1. On the y-axis, 5x − 2y + 6 = 0 gives Q = (0, 3).
2. The length of the tangent from Q to the circle is \( \sqrt{S_{11}}=\sqrt{0+9+0+18-2}=\sqrt{25}=5 \).

Teacher's Note:
1. Find Q first.
2. PQ is the length of the tangent from Q to the circle.

Answer: (c) x² + y² − 2x + 2y = 47

Explanation:
1. The diameters meet at the centre: solving gives (1, −1).
2. Area \( \pi r^2=154\Rightarrow r^2=49 \).
3. Equation: \( (x-1)^2+(y+1)^2=49\Rightarrow x^2+y^2-2x+2y=47 \).

Teacher's Note:
1. The centre is the intersection of the two diameters.
2. Use the area to find r².

Answer: (d) \( (-\sqrt2,-1)\cup(1,\sqrt2) \)

Explanation:
1. \( \sin\frac\theta2=\frac1{|a|} \) (radius over distance from the centre).
2. For \( \frac\pi2<\theta<\pi \): \( \frac{\pi}4<\frac\theta2<\frac\pi2 \), so \( \frac1{\sqrt2}<\frac1{|a|}<1 \).
3. Hence \( 1<|a|<\sqrt2 \), i.e. \( a\in(-\sqrt2,-1)\cup(1,\sqrt2) \).

Teacher's Note:
1. The half-angle between the tangents satisfies sin(θ/2) = r/d.
2. Remember both signs of a.

Answer: (a) x² + y² − 6x − 8y − 25 = 0

Explanation:
1. The circle has centre (3, 4) and radius 5. The locus is its director circle, with the same centre and radius \( 5\sqrt2 \).
2. \( (x-3)^2+(y-4)^2=50\Rightarrow x^2+y^2-6x-8y-25=0 \).

Teacher's Note:
1. Perpendicular tangents come from the director circle: radius √2 times the original.
2. Expand to compare with the options.

Answer: (b) a circle

Explanation:
1. Let \( X=\sqrt{3h+2},\ Y=\sqrt{3k} \). Then \( X^2=3h+2 \) and \( Y^2=3k \).
2. Adding: \( X^2+Y^2=3(h+k)+2=3(1)+2=5 \): a circle.

Teacher's Note:
1. Express h and k in terms of the new coordinates.
2. Use h + k = 1.

Answer: (a) \( \left(-\frac74,\frac54\right) \)

Explanation:
1. The points are \( \left(-\frac12,0\right),\ (0,1) \) (from 2x − y + 1 = 0) and \( (-3,0),\ \left(0,\frac32\right) \) (from x − 2y + 3 = 0).
2. The centre lies on the perpendicular bisectors of the two x-axis points and of the two y-axis points: \( x=\frac{-\frac12-3}{2}=-\frac74 \) and \( y=\frac{1+\frac32}{2}=\frac54 \).

Teacher's Note:
1. A chord along an axis has its perpendicular bisector parallel to the other axis.
2. The centre is at the mean of the intercepts.

Answer: (b) \( \left(\frac3{25},\frac4{25}\right) \)

Explanation:
1. Members of the system containing the point circle \( x^2+y^2=0 \) are \( (1+\lambda)(x^2+y^2)-6x-8y+1=0 \).
2. Let \( \mu=1+\lambda \): centre \( \left(\frac3\mu,\frac4\mu\right) \) and radius² \( \frac{25}{\mu^2}-\frac1\mu \).
3. A point circle needs radius 0: \( \mu=25 \). The other limiting point is \( \left(\frac3{25},\frac4{25}\right) \).

Teacher's Note:
1. Write the general member of the coaxal system with the origin as a point circle.
2. Set the radius to zero to get the limiting points.

Answer: (b) 2

Explanation:
1. The circle has centre (7, 5) and radius \( \sqrt{49+25+151}=15 \).
2. The distance from (2, −7) to the centre is \( \sqrt{25+144}=13<15 \), so the point is inside the circle.
3. The shortest distance to the circle is \( 15-13=2 \).

Teacher's Note:
1. Compare the distance from the point to the centre with the radius.
2. For an interior point the shortest distance is r − d.

Answer: (d) (x − 17)² + (y − 16)² = 1

Explanation:
1. The image of the centre (3, 2) in the line x + y = 19 is \( (19-2,\ 19-3)=(17,16) \).
2. The radius stays 1, so the image circle is \( (x-17)^2+(y-16)^2=1 \).

Teacher's Note:
1. Reflection in x + y = c sends (x, y) to (c − y, c − x).
2. The radius is unchanged.

Answer: (d) y² − 10x − 6y + 14 = 0

Explanation:
1. The given circle has centre (3, 3) and radius \( \sqrt{9+9-14}=2 \).
2. Let the moving centre be (h, k); touching the y-axis gives radius h, and touching externally gives \( \sqrt{(h-3)^2+(k-3)^2}=h+2 \).
3. Squaring: \( k^2-6k-10h+14=0 \). The locus is \( y^2-10x-6y+14=0 \).

Teacher's Note:
1. Radius of the moving circle = |h| as it touches the y-axis.
2. External contact: distance between centres = sum of radii.