JEE Mathematics Circles MCQs Set 14

Answer: (a) \( 3x-2y=3 \)

Explanation:
1. The centre is \( \left(\frac13,-1\right) \).
2. A line perpendicular to \( 2x+3y=12 \) has the form \( 3x-2y=c \); through the centre: \( c=1+2=3 \).

Teacher's Note:
1. Perpendicular lines swap the coefficients with a sign change.
2. The diameter passes through the centre.

Answer: (d) \( \left(-\frac ln,-\frac mn\right) \)

Explanation:
1. The polar of \( (x_1,y_1) \) is \( xx_1+yy_1=1 \). Compare with \( lx+my=-n \).
2. \( \frac{x_1}l=\frac{y_1}m=\frac1{-n} \), so the pole is \( \left(-\frac ln,-\frac mn\right) \).

Teacher's Note:
1. Write the line as lx + my = −n.
2. Compare coefficients with the polar.

Answer: (a) \( \left(\frac{a^2A}{-C},\frac{a^2B}{-C}\right) \)

Explanation:
1. The polar of \( (x',y') \) is \( x'x+y'y=a^2 \). Compare with \( Ax+By=-C \).
2. \( \frac{x'}A=\frac{y'}B=\frac{a^2}{-C} \), so \( (x',y')=\left(-\frac{a^2A}C,-\frac{a^2B}C\right) \).

Teacher's Note:
1. Write the line as Ax + By = −C.
2. Equate ratios of the coefficients.

Answer: (c) \( c^2=r^2(\lambda^2+\mu^2) \)

Explanation:
1. The polar of the origin is \( \lambda x+\mu y+c=0 \).
2. It touches \( x^2+y^2=r^2 \) when \( \frac{|c|}{\sqrt{\lambda^2+\mu^2}}=r \), i.e. \( c^2=r^2(\lambda^2+\mu^2) \).

Teacher's Note:
1. T = 0 at the origin gives the polar.
2. Tangent: distance from the centre = radius.

Answer: (a) (32, 48)

Explanation:
1. The polar of \( (x_1,y_1) \) is \( xx_1+yy_1=64 \). Compare with \( 2x+3y=4 \).
2. \( \frac{x_1}2=\frac{y_1}3=\frac{64}4=16 \), so the pole is \( (32,48) \).

Teacher's Note:
1. Equate ratios of coefficients.
2. The common ratio is 64/4.

Answer: (d) Intersecting at a point

Explanation:
1. Polar with respect to \( S_1 \): \( -x+2y+3(y+2)+7=0 \), i.e. \( -x+5y+13=0 \).
2. Polar with respect to \( S_2 \): \( -x+2y+3(x-1)+1=0 \), i.e. \( x+y-1=0 \).
3. Slopes \( \frac15 \) and \( -1 \): not equal (not parallel), product not \( -1 \) (not perpendicular), and the lines are different, so they intersect at a point.

Teacher's Note:
1. Write both polars using T = 0.
2. Compare their slopes.

Answer: (a) (16, 12)

Explanation:
1. \( \frac{x_1}4=\frac{y_1}3=\frac{200}{50}=4 \).
2. The pole is \( (16,12) \).

Teacher's Note:
1. Compare xx₁ + yy₁ = 200 with 4x + 3y = 50.
2. The common ratio is 4.

Answer: (a) \( x=0 \)

Explanation:
1. \( T=0 \): \( -2x+3y-2(x-2)-3(y+3)+5=0 \).
2. Simplifying: \( -4x=0 \), so \( x=0 \).

Teacher's Note:
1. Use xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
2. The y terms cancel.

Answer: (b) \( x^2+y^2=\frac{c^2}9 \)

Explanation:
1. If the pole is at distance \( d=3c \) from the centre, its polar is at distance \( \frac{c^2}d=\frac c3 \) from the centre.
2. So the line touches the circle of radius \( \frac c3 \): \( x^2+y^2=\frac{c^2}9 \).

Teacher's Note:
1. The polar lies at distance r²/d from the centre.
2. d = 3c.

Answer: (a) The pole of the line \( x-2y+5=0 \) is \( (1,1) \)

Explanation:
1. The polar of \( (1,1) \) is \( x+y-2(x+1)+(y+1)-4=0 \), i.e. \( -x+2y-5=0 \), which is \( x-2y+5=0 \). So \( (1,1) \) is the pole of that line (option a true, option c false).
2. \( S_1=1+1-4+2-4=-4<0 \), so \( (1,1) \) is inside the circle and there are no real tangents from it (option b false).

Teacher's Note:
1. Find the polar of (1, 1) using T = 0.
2. A chord of contact needs an external point.

Answer: (a) (4, 3) and (4, -3)

Explanation:
1. Subtracting: \( 8x-32=0 \), so \( x=4 \).
2. Then \( y^2=25-16=9 \), so \( y=\pm3 \).

Teacher's Note:
1. Subtract the equations to get the common chord.
2. Substitute back.

Answer: (d) They touch each other

Explanation:
1. Centres \( (3,1),\ (-1,4) \); radii \( 3 \) and \( 2 \).
2. \( d=\sqrt{16+9}=5=3+2 \), so they touch externally.

Teacher's Note:
1. Compute d, r₁ and r₂.
2. d = r₁ + r₂ means external contact.

Answer: (a) Touches

Explanation:
1. The tangent at \( (1,-2) \) is \( x-2y=5 \).
2. The second circle has centre \( (4,-3) \) and \( r=\sqrt{16+9-20}=\sqrt5 \).
3. Distance of the centre from the line: \( \frac{|4+6-5|}{\sqrt5}=\sqrt5=r \), so it touches.

Teacher's Note:
1. Find the tangent first with T = 0.
2. Compare the centre's distance with the radius.

Answer: (a) \( 23x^2+23y^2-156x+38y+168=0 \)

Explanation:
1. Family: \( S_1+\lambda S_2=0 \). Through \( (3,-3) \): \( S_1=9+9-24+6+7=7 \), \( S_2=9+9-12-30+8=-16 \), so \( 7-16\lambda=0 \), \( \lambda=\frac7{16} \).
2. \( 16S_1+7S_2=0 \): \( 23x^2+23y^2-156x+38y+168=0 \).

Teacher's Note:
1. Use S₁ + λS₂ = 0.
2. Find λ from the given point.

Answer: (d) \( y^2-10x-6y+14=0 \)

Explanation:
1. The given circle has centre \( (3,3) \) and \( r=2 \). Let the centre be \( (h,k) \) with radius \( h \) (touching the \( y\text{-axis} \)).
2. External contact: \( (h-3)^2+(k-3)^2=(h+2)^2 \).
3. This gives \( k^2-6k-10h+14=0 \). The locus is \( y^2-10x-6y+14=0 \).

Teacher's Note:
1. Radius equals the distance from the y-axis.
2. Distance between centres = sum of radii.

Answer: (a) \( S+\lambda P=0 \)

Explanation:
1. The standard family is \( S+\lambda P=0 \).

Teacher's Note:
1. It is the family of circles through the intersection points.
2. The coefficient of S is kept 1 so that it remains a circle.

Answer: (b) One of these circles lies entirely inside the other

Explanation:
1. Centres \( (0,1),\ (1,0) \); radii 3 and 5; \( d=\sqrt2 \).
2. \( r_2-r_1=2>\sqrt2=d \), so the smaller circle lies entirely inside the larger.

Teacher's Note:
1. Compare d with the difference of radii.
2. d < r₂ − r₁ means one circle is inside the other.

Answer: (c) 16

Explanation:
1. The second circle has centre \( (5,0) \) and \( r=\sqrt{25-\lambda} \).
2. External contact: \( 5=2+\sqrt{25-\lambda} \), so \( 25-\lambda=9 \) and \( \lambda=16 \).

Teacher's Note:
1. Distance between centres = sum of radii.
2. Square to remove the root.

Answer: (c) \( 7x^2+7y^2-10x-10y-12=0 \)

Explanation:
1. Family \( S_1+\lambda S_2=0 \) has centre \( \left(\frac{6-2\lambda}{2(1+\lambda)},\frac{-2+4\lambda}{2(1+\lambda)}\right) \).
2. Centre on \( y=x \): \( 6-2\lambda=-2+4\lambda \), so \( \lambda=\frac43 \).
3. \( 3S_1+4S_2=0 \): \( 7x^2+7y^2-10x-10y-12=0 \).

Teacher's Note:
1. Write the centre of the family in terms of λ.
2. Equate its coordinates.

Answer: (b) \( x^2+y^2-3x+1=0 \)

Explanation:
1. Family: \( x^2+y^2-6x+8+\lambda(x^2+y^2-6)=0 \). At \( (1,1) \): \( 4-4\lambda=0 \), so \( \lambda=1 \).
2. \( 2x^2+2y^2-6x+2=0 \), i.e. \( x^2+y^2-3x+1=0 \).

Teacher's Note:
1. Use S₁ + λS₂ = 0.
2. Substitute (1, 1).

Answer: (a) \( x^2+y^2-6x+7=0 \)

Explanation:
1. Family \( S_1+\lambda S_2=0 \) has centre \( \left(\frac{1+2\lambda}{1+\lambda},\frac{2+\lambda}{1+\lambda}\right) \).
2. On \( x+2y-3=0 \): \( (1+2\lambda)+2(2+\lambda)=3(1+\lambda) \), so \( 5+4\lambda=3+3\lambda \) and \( \lambda=-2 \).
3. \( S_1-2S_2=0 \) gives \( -x^2-y^2+6x-7=0 \), i.e. \( x^2+y^2-6x+7=0 \).

Teacher's Note:
1. Write the centre of the family in terms of λ.
2. Substitute it in the given line.