JEE Mathematics Circles MCQs Set 13

Answer: (c) \( 4x-3y=0 \)

Explanation:
1. The centre is \( (-3,-4) \).
2. The normal through the origin and the centre has slope \( \frac43 \): \( 4x-3y=0 \).

Teacher's Note:
1. The normal passes through the centre.
2. Use the two points (0,0) and (−3, −4).

Answer: (a) \( x+4y=0 \)

Explanation:
1. The centre is \( (20,-5) \).
2. The normal through \( (4,-1) \) and the centre has slope \( \frac{-5+1}{20-4}=-\frac14 \): \( y+1=-\frac14(x-4) \), i.e. \( x+4y=0 \).

Teacher's Note:
1. Every normal passes through the centre.
2. Find the slope from the two points.

Answer: (a) \( 2\sqrt3 \)

Explanation:
1. The tangent is \( x+\sqrt3y=4 \), meeting the \( x\text{-axis} \) at \( (4,0) \). The normal passes through the origin.
2. The triangle has vertices \( (0,0),\ (4,0),\ (1,\sqrt3) \).
3. Area \( =\frac12\cdot4\cdot\sqrt3=2\sqrt3 \).

Teacher's Note:
1. The normal at any point of a circle passes through the centre.
2. Use base along the x-axis and the height √3.

Answer: (c) \( -\frac43,1 \)

Explanation:
1. Normal to the second circle (centre \( (0,2) \)): \( 2b=2 \), so \( b=1 \).
2. Tangent to the first (centre \( (1,0) \), \( r=2 \)): \( \frac{|a-2|}{\sqrt{a^2+1}}=2 \) gives \( 3a^2+4a=0 \).
3. \( a\neq0 \), so \( a=-\frac43 \).

Teacher's Note:
1. A normal passes through the centre; a tangent is at distance r from the centre.
2. Use the normal condition first to get b.

Answer: (b) \( a\in[2,3],\ b\in[-3,2] \)

Explanation:
1. The point lies on the segment between the ends, so its coordinates lie between theirs.
2. \( a \) lies between 2 and 3; \( b \) lies between \( -3 \) and 2.

Teacher's Note:
1. The chord is a line segment.
2. Compare x and y ranges separately.

Answer: (c) 3

Explanation:
1. The given circle has centre \( (1,3) \) and \( r^2=1+9-6=4 \), so its diameter has half-length 2.
2. The distance between the centres is \( \sqrt{1+4}=\sqrt5 \).
3. The required radius: \( R^2=2^2+(\sqrt5)^2=9 \), so \( R=3 \).

Teacher's Note:
1. The line from the new centre to the midpoint of a chord is perpendicular to the chord.
2. Use Pythagoras.

Answer: (c) \( (-\sqrt2,\sqrt2) \)

Explanation:
1. \( P \) is at angle \( 45^\circ \). A quarter circumference corresponds to \( 90^\circ \).
2. \( Q \) in the second quadrant is at \( 45^\circ+90^\circ=135^\circ \): \( (2\cos135^\circ,2\sin135^\circ)=(-\sqrt2,\sqrt2) \).

Teacher's Note:
1. A quarter of the circumference subtends 90° at the centre.
2. Rotate P by 90° anticlockwise.

Answer: (a) 10

Explanation:
1. The line is \( y-\sqrt8=-(x+\sqrt8) \), i.e. \( x+y=0 \), which passes through the centre.
2. So \( AB \) is a diameter: \( AB=2\cdot5=10 \).

Teacher's Note:
1. Find the line's equation first.
2. A line through the centre cuts a diameter.

Answer: (c) \( |a|<4 \)

Explanation:
1. The chord is \( x+y=a \). It cuts the circle when its distance from the origin is less than the radius: \( \frac{|a|}{\sqrt2}<2\sqrt2 \).
2. So \( |a|<4 \).

Teacher's Note:
1. Equal intercepts give the line x + y = a.
2. The distance from the centre must be less than r.

Answer: (c) (-25, 50)

Explanation:
1. The required point is the pole of the line. Its polar is \( xx_1+yy_1=25 \).
2. Compare \( xx_1+yy_1-25=0 \) with \( x-2y+1=0 \): \( \frac{x_1}1=\frac{y_1}{-2}=\frac{-25}1 \).
3. So \( x_1=-25 \) and \( y_1=50 \).

Teacher's Note:
1. The tangents at the ends of a chord meet at the pole of the chord.
2. Match coefficients of the polar with the given line.

Answer: (b) \( \left(\frac12,\frac14\right) \)

Explanation:
1. For \( (x_1,y_1) \) on the line, \( y_1=4-2x_1 \). The chord of contact is \( xx_1+yy_1=1 \).
2. Substituting: \( x_1(x-2y)+4y-1=0 \), a family through the intersection of \( x-2y=0 \) and \( 4y-1=0 \).
3. That point is \( \left(\frac12,\frac14\right) \).

Teacher's Note:
1. Express the chord of contact with one parameter.
2. A family a·u + b·v = 0 passes through the point where u = v = 0.

Answer: (a) \( \left(\frac85\sqrt6,\frac45\right) \)

Explanation:
1. Tangent from \( (0,5) \): \( y=mx+5 \) with \( \frac5{\sqrt{1+m^2}}=1 \), so \( m=\pm2\sqrt6 \).
2. \( C \) is the pole of the chord \( mx-y+5=0 \) with respect to \( x^2+y^2=4 \): \( \frac{x_1}m=\frac{y_1}{-1}=\frac4{-5} \), so \( C=\left(-\frac{4m}5,\frac45\right) \).
3. For \( m=-2\sqrt6 \), \( C=\left(\frac{8\sqrt6}5,\frac45\right) \), which is option (a). The \( y \)-coordinate is \( +\frac45 \), which rules out (b) and (c).

Teacher's Note:
1. The tangents at A and B meet at the pole of AB.
2. Use the pole formula for x² + y² = 4.

Answer: (c) \( xx_1+yy_1=x_1^2+y_1^2 \)

Explanation:
1. The chord with given midpoint is \( T=S_1 \).
2. \( xx_1+yy_1-a^2=x_1^2+y_1^2-a^2 \), i.e. \( xx_1+yy_1=x_1^2+y_1^2 \).

Teacher's Note:
1. Use T = S₁.
2. Option (d) is the chord of contact, not the midpoint chord.

Answer: (c) \( x^2+y^2-x=0 \)

Explanation:
1. The midpoint \( M \) of a chord \( OP \) satisfies \( CM\perp OM \) where \( C=(1,0) \).
2. So \( M \) lies on the circle with diameter \( OC \): \( x(x-1)+y\cdot y=0 \), i.e. \( x^2+y^2-x=0 \).

Teacher's Note:
1. The line from the centre to a chord's midpoint is perpendicular to the chord.
2. The locus is the circle on OC as diameter.

Answer: (a) \( x^2+y^2-hx-ky=0 \)

Explanation:
1. For the midpoint \( M(x,y) \), \( T=S_1 \) gives \( xx_1+yy_1=x^2+y^2 \) with the chord through \( (h,k) \): \( hx+ky=x^2+y^2 \).
2. So the locus is \( x^2+y^2-hx-ky=0 \).

Teacher's Note:
1. Substitute the fixed point into the chord T = S₁.
2. The locus is the circle on the segment from the centre to (h, k) as diameter.

Answer: (d) \( x-y=0 \)

Explanation:
1. \( T=S_1 \): \( x+y-2(x+1)=1+1-4 \).
2. \( -x+y-2=-2 \), so \( x-y=0 \).

Teacher's Note:
1. Use T = S₁.
2. The chord is perpendicular to the line from the centre (2, 0) to (1, 1).

Answer: (d) \( x^2+y^2-x-3y=0 \)

Explanation:
1. The centre is \( C(1,3) \). The midpoint lies on the circle with diameter \( OC \).
2. \( x(x-1)+y(y-3)=0 \), i.e. \( x^2+y^2-x-3y=0 \).

Teacher's Note:
1. The origin lies inside the circle, so every line through it is a chord.
2. Locus = circle on OC as diameter.

Answer: (c) \( 6\sqrt3 \)

Explanation:
1. The distance of the origin from the line is \( \frac{15}5=3 \).
2. Chord length \( =2\sqrt{36-9}=2\sqrt{27}=6\sqrt3 \).

Teacher's Note:
1. Chord length = 2√(r² − d²).
2. Find d with the point-to-line distance.

Answer: (a) \( x+y=2 \)

Explanation:
1. For midpoint \( (h,k) \), the chord is \( T=S_1 \): \( (h-1)x+(k-1)y=h^2+k^2-h-k=c \).
2. Homogenise the circle with it: \( x^2+y^2-\frac{2(x+y)((h-1)x+(k-1)y)}{c}=0 \).
3. Equally inclined lines (slopes \( m,-m \)) mean no \( xy \) term: \( (h-1)+(k-1)=0 \), so \( h+k=2 \). The locus is \( x+y=2 \).

Teacher's Note:
1. Join the origin to the intersection points by homogenising the circle with the chord.
2. Lines equally inclined to the x-axis have slopes m and −m, so the xy term vanishes.

Answer: (d) \( x^2+y^2=a^2-l^2 \)

Explanation:
1. If \( M \) is the midpoint, then \( OM^2=a^2-l^2 \) (right triangle with half-chord \( l \)).
2. So the locus is \( x^2+y^2=a^2-l^2 \).

Teacher's Note:
1. The half chord, the radius and OM form a right triangle.
2. OM² = a² − l².

Answer: (a) \( \frac12 \)

Explanation:
1. The centre is \( \left(-\frac34,-1\right) \).
2. It lies on the line: \( -1=-\frac32+k \), so \( k=\frac12 \).

Teacher's Note:
1. A diameter passes through the centre.
2. Divide by 2 to find the centre.

Answer: (c) A concentric circle of radius \( \sqrt2\,a \)

Explanation:
1. The point of intersection of perpendicular tangents lies at distance \( \sqrt2a \) from the centre.
2. The locus is the director circle \( x^2+y^2=2a^2 \).

Teacher's Note:
1. The two radii and the tangents form a square.
2. Its diagonal is a√2.