JEE Mathematics Circles MCQs Set 15

Answer: (b) \( (0,0),\ \left(\frac{2ab^2}{a^2+b^2},\frac{2ba^2}{a^2+b^2}\right) \)

Explanation:
1. Subtracting: \( ax=by \), so \( y=\frac{a}{b}x \).
2. Substituting: \( x^2\left(1+\frac{a^2}{b^2}\right)=2ax \), so \( x=0 \) or \( x=\frac{2ab^2}{a^2+b^2} \), giving \( y=\frac{2a^2b}{a^2+b^2} \).

Teacher's Note:
1. The common chord is the difference of the two equations.
2. Both circles pass through the origin.

Answer: (c) Touch internally

Explanation:
1. Centres \( (1,-3),\ \left(\frac52,-3\right) \); radii \( 2 \) and \( \frac12 \).
2. \( d=\frac32=2-\frac12 \), so they touch internally.

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

Answer: (c) \( \frac1{a^2}+\frac1{b^2}=\frac1{c^2} \)

Explanation:
1. Centres \( (a,0),\ (0,b) \); radii \( \sqrt{a^2-c^2},\ \sqrt{b^2-c^2} \).
2. External contact: \( a^2+b^2=r_1^2+r_2^2+2r_1r_2 \) gives \( 2c^2=2r_1r_2 \), so \( c^4=(a^2-c^2)(b^2-c^2) \).
3. This reduces to \( c^2(a^2+b^2)=a^2b^2 \), i.e. \( \frac1{a^2}+\frac1{b^2}=\frac1{c^2} \).

Teacher's Note:
1. Square the distance condition d = r₁ + r₂.
2. Divide by a²b²c² at the end.

Answer: (b) Touch each other internally

Explanation:
1. Centres \( (0,0),\ (1,2) \); radii \( \sqrt5 \) and \( 2\sqrt5 \).
2. \( d=\sqrt5=2\sqrt5-\sqrt5 \), so they touch internally.

Teacher's Note:
1. Compute the radii and d.
2. d = r₂ − r₁.

Answer: (b) They cut each other orthogonally

Explanation:
1. Centres \( (0,13),\ (0,0) \); radii \( 12 \) and \( 5 \); \( d=13 \).
2. \( d^2=169=12^2+5^2 \), so the circles cut orthogonally.

Teacher's Note:
1. Orthogonality: d² = r₁² + r₂².
2. Check 144 + 25 = 169.

Answer: (a) \( x^2+y^2-4x-2y-20=0 \)

Explanation:
1. \( C_1 \) has centre \( (2,1) \) and radius \( 4 \).
2. The centre of \( C_2 \) stays at distance \( 4+1=5 \) from \( (2,1) \): \( (x-2)^2+(y-1)^2=25 \).
3. That is \( x^2+y^2-4x-2y-20=0 \).

Teacher's Note:
1. Rolling outside keeps the circles externally tangent.
2. Locus radius = r₁ + r₂.

Answer: (a) 15

Explanation:
1. The second circle has centre \( (0,-4) \) and radius \( \sqrt{16-c} \).
2. External contact: \( 4=3+\sqrt{16-c} \), so \( 16-c=1 \) and \( c=15 \).

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

Answer: (c) \( 2<r<8 \)

Explanation:
1. The first circle has centre \( (5,0) \) and radius \( 3 \); the distance between centres is \( 5 \).
2. Two intersection points need \( |r-3|<5<r+3 \), so \( r>2 \) and \( r<8 \).

Teacher's Note:
1. Use |r₁ − r₂| < d < r₁ + r₂.
2. Equality gives tangency, not two points.

Answer: (b) \( x^2+y^2+gx+fy=0 \)

Explanation:
1. The angles at \( P \) and \( Q \) are right angles, so \( OC \) is a diameter of the circumcircle, where \( C=(-g,-f) \).
2. The circle on \( OC \) as diameter: \( x(x+g)+y(y+f)=0 \), i.e. \( x^2+y^2+gx+fy=0 \).

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact.
2. The hypotenuse OC is the diameter.

Answer: (b) \( x^2+y^2-x-y-8=0 \)

Explanation:
1. The smallest circle has the chord as diameter. Family: \( x^2+y^2-9+\lambda(x+y-1)=0 \) with centre \( \left(-\frac\lambda2,-\frac\lambda2\right) \).
2. The centre lies on the line: \( -\lambda=1 \), so \( \lambda=-1 \).
3. \( x^2+y^2-x-y-8=0 \).

Teacher's Note:
1. Smallest circle through two points = circle on the segment as diameter.
2. The centre must lie on the chord.

Answer: (c) 3

Explanation:
1. Centres \( (2,3),\ (-3,-9) \); radii \( 5 \) and \( 8 \).
2. \( d=\sqrt{25+144}=13=5+8 \), so they touch externally: 3 common tangents.

Teacher's Note:
1. Compute d and compare with r₁ + r₂.
2. External contact gives 3 common tangents.

Answer: (c) 3

Explanation:
1. Centres \( (0,0),\ (4,0) \); radii \( 2 \) and \( 2 \).
2. \( d=4=2+2 \), so they touch externally: 3 common tangents.

Teacher's Note:
1. Check d against r₁ + r₂.
2. Touching externally: 2 direct + 1 transverse tangent.

Answer: (b) \( y=3 \)

Explanation:
1. The second circle has centre \( (0,6) \) and radius \( 3 \), so the circles touch externally at \( (0,3) \).
2. The common tangent at the point of contact is perpendicular to the line of centres (the \( y\text{-axis} \)): \( y=3 \).

Teacher's Note:
1. Find the point of contact first.
2. The tangent is perpendicular to the line of centres.

Answer: (c) 3

Explanation:
1. Centres \( (-1,-4),\ (2,5) \); radii \( 2\sqrt{10} \) and \( \sqrt{10} \).
2. \( d=\sqrt{90}=3\sqrt{10}=r_1+r_2 \), so they touch externally: 3 common tangents.

Teacher's Note:
1. Compute radii: r₁² = 40, r₂² = 10.
2. d = r₁ + r₂.

Answer: (c) (4, 5), (-2, -3)

Explanation:
1. The centres lie on the normal at \( (1,1) \), at distance 5 on either side. The unit normal is \( \frac{(3,4)}5 \).
2. Centres \( =(1,1)\pm5\cdot\frac{(3,4)}5=(4,5)\text{ and }(-2,-3) \).

Teacher's Note:
1. The radius at the point of contact is perpendicular to the tangent.
2. Move 5 units along the unit normal in both directions.

Answer: (b) \( \lambda>\frac{13}8 \) or \( \lambda<-\frac{13}8 \)

Explanation:
1. The second circle has centre \( (2\lambda,0) \) and radius \( \sqrt{4\lambda^2-9} \) (real for \( |\lambda|>\frac32 \)).
2. Two common tangents mean the circles intersect in two points: \( d<r_1+r_2 \), i.e. \( 2|\lambda|-2<\sqrt{4\lambda^2-9} \), which gives \( |\lambda|>\frac{13}8 \).
3. The condition \( d>|r_2-r_1| \) then holds automatically.

Teacher's Note:
1. Two common tangents means two intersection points.
2. Square the inequality carefully.

Answer: (a) \( x+y+1=0 \)

Explanation:
1. The common chord is \( S_1-S_2=0 \): \( -2x-2y-2=0 \).
2. So \( x+y+1=0 \).

Teacher's Note:
1. Subtract the two equations.
2. Divide by −2.

Answer: (a) \( 10x-3y-18=0 \)

Explanation:
1. Divide the first by 3: \( x^2+y^2-\frac23x+4y-3=0 \).
2. Subtract the second: \( -\frac{20}3x+2y+12=0 \), i.e. \( 10x-3y-18=0 \).

Teacher's Note:
1. First make the coefficients of x² and y² equal to 1.
2. Then subtract.

Answer: (b) \( 2\sqrt2 \)

Explanation:
1. The common chord is \( -2x-1=0 \), i.e. \( x=-\frac12 \).
2. Substituting in the first: \( y^2+3y+\frac14=0 \), so \( |y_1-y_2|=\sqrt{9-1}=2\sqrt2 \).

Teacher's Note:
1. The chord is a vertical line.
2. Length = difference of the two y roots.

Answer: (b) \( 2\sqrt{g'^2+f'^2-c'} \)

Explanation:
1. The common chord is a diameter of the second circle.
2. Its length is \( 2r'=2\sqrt{g'^2+f'^2-c'} \).

Teacher's Note:
1. Bisecting means the chord passes through the centre of the second circle.
2. Diameter = 2r.

Answer: (a) 2 units

Explanation:
1. The common chord is \( S_1-S_2=0 \): \( -12x-5y-4=0 \), i.e. \( 12x+5y+4=0 \).
2. Distance \( =\frac{|12+10+4|}{13}=2 \).

Teacher's Note:
1. Subtract the equations to get the chord.
2. Use the point-to-line distance formula.

Answer: (a) \( 4\sqrt2 \)

Explanation:
1. The common chord is \( 4x-3y-10=0 \). Its distance from the origin is \( \frac{10}5=2 \).
2. Length \( =2\sqrt{12-4}=2\sqrt8=4\sqrt2 \).

Teacher's Note:
1. Find the chord, then its distance from the centre of x² + y² = 12.
2. Length = 2√(r² − d²).