JEE Mathematics Circle and Conic Section MCQs Set 26

Answer: (c) \( c>0 \)

Explanation:
1. Subtracting the equations gives \( (\lambda_1-\lambda_2)(x-y)=0 \), so any common point lies on \( x=y \), and there \( 2x^2+c=0 \).
2. If \( c<0 \) this has two real roots, so the circles intersect; if \( c=0 \) they touch at the origin.
3. One circle lies inside the other only when there is no common point: \( 2x^2=-c \) has no real solution, i.e. \( c>0 \).

Teacher's Note:
1. Two circles are nested exactly when they have no point in common and are not separate.
2. Find the common points by subtracting the circle equations.

Answer: (b) \( \frac{b}{a-c}>2 \)

Explanation:
1. A normal of slope \( m \) to \( y^2=4ax \) is \( y=mx-2am-am^3 \), and to \( y^2=4c(x-b) \) it is \( y=m(x-b)-2cm-cm^3 \).
2. Equating intercepts for \( m\ne0 \): \( -2a-am^2=-b-2c-cm^2 \), so \( m^2(c-a)=2a-2c-b \).
3. \( m^2=\dfrac{2(a-c)-b}{c-a}=\dfrac{b}{a-c}-2 \). A real \( m \) requires \( m^2>0 \), so \( \dfrac{b}{a-c}>2 \).

Teacher's Note:
1. A normal of slope m to \( y^2=4ax \) has the form \( y=mx-2am-am^3 \); learn this standard form.
2. The shifted parabola uses \( x-b \) in place of \( x \).

Answer: (d) \( \frac{25}{3} \)

Explanation:
1. The normal at \( (2\sec\theta,\tan\theta) \) is \( 2x\cos\theta+y\cot\theta=5 \), with intercepts \( \dfrac{5\sec\theta}2 \) and \( 5\tan\theta \).
2. Equal intercepts give \( \sin\theta=\tfrac12 \), so \( \theta=30^\circ \) and the normal is \( \sqrt3x+\sqrt3y=5 \), i.e. \( x+y=\dfrac5{\sqrt3} \).
3. The line \( x+y=k \) touches the ellipse when \( k^2=a^2+b^2 \). So \( a^2+b^2=\dfrac{25}{3} \).

Teacher's Note:
1. For the line \( lx+my=n \) to touch \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \), \( n^2=a^2l^2+b^2m^2 \).
2. Use the parametric form of a point on the hyperbola to write the normal.

Answer: (a) \( K\in(0,2) \)

Explanation:
1. The left side is the absolute difference of the distances of \( (x,y) \) from the foci \( (0,1) \) and \( (0,-1) \), which are 2 apart.
2. This locus is a hyperbola with \( 2a=K \) and \( 2c=2 \), which requires \( 0<2a<2c \).
3. So \( 0<K<2 \).

Teacher's Note:
1. A hyperbola needs the constant difference to be less than the distance between the foci.
2. At \( K=2 \) the locus degenerates to two rays and at \( K=0 \) to a line.

Answer: (a) circle

Explanation:
1. \( A=(8,0) \), \( B=(0,6) \); \( C=(6,0) \), \( D=(0,8) \).
2. \( A \) and \( C \) are on the \( x \)-axis and \( B,D \) on the \( y \)-axis with \( OA\cdot OC=48=OB\cdot OD \).
3. By the converse of the intersecting-chords theorem, the four points are concyclic.

Teacher's Note:
1. If two lines meet at O and \( OA\cdot OC=OB\cdot OD \) (same side), the four points lie on a circle.
2. You can also verify that a circle through A, B, C passes through D.

Answer: (b) 2

Explanation:
1. The centre lies on the perpendicular bisectors of the two chords: \( \left(\dfrac{1+t}2,\dfrac{1+t}2\right) \), with radius squared \( \dfrac{(1-t)^2}{2} \).
2. The circle: \( x^2+y^2-(1+t)(x+y)+2t=0 \), i.e. \( (x^2+y^2-x-y)+t(2-x-y)=0 \).
3. For this to hold for all \( t \): \( x+y=2 \) and \( x^2+y^2=2 \), so \( (a,b)=(1,1) \) and \( a^2+b^2=2 \).

Teacher's Note:
1. A family of curves \( S+tL=0 \) all pass through the common points of \( S=0 \) and \( L=0 \).
2. Separate the equation into the part with \( t \) and the part without.

Answer: (d) 5

Explanation:
1. Take \( A=(2\sec\theta,\tan\theta) \); the tangent is \( \dfrac{x\sec\theta}2-y\tan\theta=1 \).
2. Meeting \( y=\pm\dfrac x2 \) gives \( x=\dfrac{2}{\sec\theta\mp\tan\theta} \) so \( CQ=\sqrt5\cdot\dfrac{1}{\sec\theta-\tan\theta} \) and \( CR=\sqrt5\cdot\dfrac1{\sec\theta+\tan\theta} \) (distances along the asymptotes).
3. \( CQ\cdot CR=\dfrac{5}{\sec^2\theta-\tan^2\theta}=5 \).

Teacher's Note:
1. The product of the distances of the tangent's intercepts on the asymptotes from the centre is the constant \( a^2+b^2 \).
2. This holds for every point of every hyperbola.

Answer: (c) 4

Explanation:
1. Points on a line through \( P \) with direction \( (\cos\theta,\sin\theta) \): \( (t+r\sin\theta)^2=4(1+r\cos\theta) \), i.e. \( r^2\sin^2\theta+(2t\sin\theta-4\cos\theta)r+t^2-4=0 \).
2. \( PA\cdot PB=|r_1r_2|=\dfrac{|t^2-4|}{\sin^2\theta}=3|t| \), so \( |t^2-4|=3|t|\sin^2\theta\le3|t| \).
3. For \( t>0 \) with \( t^2\ge4 \): \( t^2-3t-4\le0 \), so \( t\le4 \). The maximum value is 4.

Teacher's Note:
1. Product of the roots of the quadratic in the distance parameter gives the product of the distances.
2. \( \sin^2\theta\le1 \) supplies the bound.

Answer: (b) 2

Explanation:
1. The distance of the line \( x+y=n \) from the centre is \( \dfrac{n}{\sqrt2} \). A chord exists when \( \dfrac{n^2}2<4 \), i.e. \( n^2<8 \), so \( n=1,2 \).
2. Chord length squared \( =4\left(4-\dfrac{n^2}2\right)=16-2n^2 \): \( n=1 \) gives 14 and \( n=2 \) gives 8.
3. Sum \( =22=11k \), so \( k=2 \).

Teacher's Note:
1. Chord length is \( 2\sqrt{r^2-d^2} \).
2. The lines for \( n\ge3 \) do not cut the circle.

Answer: (a) \( -2\sqrt5<a<2\sqrt5 \)

Explanation:
1. The circle has centre \( (0,0) \) and radius 2.
2. Two distinct intersections need distance from the centre less than the radius: \( \dfrac{|a|}{\sqrt{1+4}}<2 \).
3. So \( |a|<2\sqrt5 \), i.e. \( -2\sqrt5<a<2\sqrt5 \).

Teacher's Note:
1. Line and circle meet in two points when the perpendicular distance from the centre is less than the radius.
2. The absolute value gives a two-sided inequality.