JEE Mathematics Circles MCQs Set 16

Answer: (d) \( \frac\pi2 \)

Explanation:
1. The common chord is \( x+y=4 \). Substituting in \( x^2+y^2=16 \): \( 2x^2-8x=0 \), so the points are \( (0,4) \) and \( (4,0) \).
2. These lie on the axes, so the angle at the origin is \( \frac\pi2 \).

Teacher's Note:
1. Subtract to find the chord, then the endpoints.
2. Compare the position vectors of the endpoints.

Answer: (b) \( \sqrt{4c^2-2(a-b)^2} \)

Explanation:
1. Subtracting the equations gives the common chord \( x=y \).
2. The distance of \( (a,b) \) from \( x=y \) is \( \frac{|a-b|}{\sqrt2} \).
3. Length \( =2\sqrt{c^2-\frac{(a-b)^2}2}=\sqrt{4c^2-2(a-b)^2} \).

Teacher's Note:
1. Find the chord first.
2. Use 2√(r² − d²).

Answer: (a) \( \left(-\frac95,\frac{12}5\right),\left(\frac95,-\frac{12}5\right) \)

Explanation:
1. The longest common chord is a diameter of \( c_1 \): the line \( y=\frac34x \) through the origin.
2. The centre of \( c_2 \) lies on the perpendicular bisector (through the origin) \( 4x+3y=0 \), at distance \( \sqrt{25-16}=3 \) from the origin.
3. Centres \( =\pm3\cdot\frac{(-3,4)}5=\pm\left(-\frac95,\frac{12}5\right) \).

Teacher's Note:
1. The centre of c₂ is on the perpendicular bisector of the common chord.
2. Use Pythagoras with half the chord (4) and radius 5.

Answer: (a) \( x^2+y^2+\frac{2ab^2}{a^2+b^2}x+\frac{2a^2b}{a^2+b^2}y+c=0 \)

Explanation:
1. The common chord is \( ax-by=0 \). Family: \( x^2+y^2+2ax+c+\lambda(ax-by)=0 \) with centre \( \left(-a-\frac{\lambda a}2,\frac{\lambda b}2\right) \).
2. On the chord: \( -a^2\left(1+\frac\lambda2\right)-\frac{\lambda b^2}2=0 \), so \( \lambda=-\frac{2a^2}{a^2+b^2} \).
3. Equation: \( x^2+y^2+\frac{2ab^2}{a^2+b^2}x+\frac{2a^2b}{a^2+b^2}y+c=0 \).

Teacher's Note:
1. The centre of the required circle lies on the chord.
2. Use S + λ(chord) = 0.

Answer: (c) \( 2x+4y-9=0 \)

Explanation:
1. Let the circle be \( x^2+y^2+2gx+2fy+c=0 \). Orthogonality with \( x^2+y^2=4 \): \( 0=c-4 \), so \( c=4 \).
2. Through \( (1,2) \): \( 5+2g+4f+4=0 \), i.e. \( 2g+4f+9=0 \).
3. With the centre \( (x,y)=(-g,-f) \): \( 2x+4y-9=0 \).

Teacher's Note:
1. Use 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
2. Replace g, f by −x, −y.

Answer: (c) \( ad+be=2c+2f \)

Explanation:
1. Here \( g_1=\frac a2,f_1=\frac b2,g_2=\frac d2,f_2=\frac e2 \).
2. \( 2g_1g_2+2f_1f_2=c_1+c_2 \) gives \( \frac{ad}2+\frac{be}2=c+f \), i.e. \( ad+be=2c+2f \).

Teacher's Note:
1. Convert to the 2g, 2f form first.
2. Apply the orthogonality condition.

Answer: (c) 3

Explanation:
1. \( d^2=(5-2)^2+(6-3)^2=18 \).
2. Orthogonal: \( d^2=2a^2 \), so \( a^2=9 \) and \( a=3 \).

Teacher's Note:
1. Orthogonal circles satisfy d² = r₁² + r₂².
2. Both radii are a.

Answer: (b) 18

Explanation:
1. Write the second as \( x^2+y^2+3x+2y+\frac c2=0 \).
2. \( 2(2)\left(\frac32\right)+2(3)(1)=3+\frac c2 \), so \( 12=3+\frac c2 \) and \( c=18 \).

Teacher's Note:
1. Divide by 2 first.
2. Use 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Answer: (a) \( 2 \) or \( -\frac32 \)

Explanation:
1. \( 2(1)(0)+2(k)(k)=6+k \), so \( 2k^2-k-6=0 \).
2. \( (k-2)(2k+3)=0 \), so \( k=2 \) or \( k=-\frac32 \).

Teacher's Note:
1. Use 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
2. Solve the quadratic.

Answer: (c) 4

Explanation:
1. Write the first as \( x^2+y^2-\frac32x+3y+\frac k2=0 \).
2. \( 2\left(-\frac34\right)(-2)+2\left(\frac32\right)(5)=\frac k2+16 \), so \( 18=\frac k2+16 \) and \( k=4 \).

Teacher's Note:
1. Divide by 2 first.
2. Apply 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Answer: (c) Cut each other orthogonally

Explanation:
1. The second is \( x^2+y^2-\frac83x+\frac{29}3y=0 \).
2. \( 2(1)\left(\frac43\right)+2(1)\left(-\frac{29}6\right)=\frac83-\frac{29}3=-7=-7+0 \), so they cut orthogonally.

Teacher's Note:
1. Divide the second equation by 3.
2. Check 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Answer: (b) \( 90^\circ \)

Explanation:
1. Centres \( (-4,1),\ (1,-4) \); \( r_1^2=16+1+9=26 \), \( r_2^2=1+16+7=24 \).
2. \( d^2=25+25=50=r_1^2+r_2^2 \), so the circles cut at \( 90^\circ \).

Teacher's Note:
1. Check d² = r₁² + r₂².
2. Compute the radii squared from g² + f² − c.

Answer: (d) \( -\frac54 \)

Explanation:
1. \( 2(\lambda)(2)+2(3)(1)=1+0 \).
2. \( 4\lambda+6=1 \), so \( \lambda=-\frac54 \).

Teacher's Note:
1. Use 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
2. Here g₁ = λ, f₁ = 3, g₂ = 2, f₂ = 1.

Answer: (a) \( x^2+y^2-2x-4y+1=0 \)

Explanation:
1. Let the circle be \( x^2+y^2+2gx+2fy+c=0 \). Orthogonality with each circle gives \( -3g-6f=c+14 \), \( -g-4f=c+8 \), \( 2g-6f=c+9 \).
2. Subtracting the first two: \( g+f=-3 \); subtracting the second from the third: \( 3g-2f=1 \). So \( g=-1 \), \( f=-2 \).
3. Then \( c=-g-4f-8=1 \). The circle is \( x^2+y^2-2x-4y+1=0 \).

Teacher's Note:
1. Write the orthogonality condition for each of the three circles.
2. Solve the three linear equations in g, f, c.

Answer: (a) 2

Explanation:
1. The family is \( x^2+y^2+\frac\lambda2x-\frac{1+\lambda^2}2y-5=0 \), so \( g_1=\frac\lambda4,\ f_1=-\frac{1+\lambda^2}4,\ c_1=-5 \).
2. Orthogonality with \( g_2=2,f_2=3,c_2=3 \): \( \lambda-\frac{3(1+\lambda^2)}2=-2 \), i.e. \( 3\lambda^2-2\lambda-1=0 \).
3. This gives \( \lambda=1 \) or \( \lambda=-\frac13 \), so 2 circles.

Teacher's Note:
1. Write the family in the standard form.
2. Count the real roots of the quadratic in λ.

Answer: (a) The orthocentre

Explanation:
1. The feet of the altitudes lie on all three circles, since each altitude foot sees the opposite side at a right angle.
2. So the altitudes are the radical axes, and they meet at the orthocentre.

Teacher's Note:
1. The radical axis of two such circles is the altitude through their common vertex.
2. Altitudes are concurrent at the orthocentre.

Answer: (d) None of these

Explanation:
1. \( S_1-S_2=0 \): \( -6y+6=0 \), so \( y=1 \).
2. \( S_1-S_3=0 \): \( 8x-4y-24=0 \), so \( x=\frac72 \).
3. The radical centre is \( \left(\frac72,1\right) \), which is none of the options.

Teacher's Note:
1. Subtract the circles pairwise.
2. Solve the two radical axes.

Answer: (d) None of these

Explanation:
1. \( S_1-S_2=0 \): \( -4x+33=0 \), so \( x=\frac{33}4 \).
2. \( S_2-S_3=0 \): \( -12x+12y+19=0 \), so \( y=\frac{33}4-\frac{19}{12}=\frac{20}3 \).
3. The radical centre is \( \left(\frac{33}4,\frac{20}3\right) \), which is none of the options.

Teacher's Note:
1. Subtract the circles pairwise.
2. Solve the two linear equations.

Answer: (b) \( x+10y-2=0 \)

Explanation:
1. The first is \( x^2+y^2-\frac73x+\frac83y+\frac{11}3=0 \).
2. Subtracting: \( \frac23x+\frac{20}3y-\frac43=0 \), i.e. \( x+10y-2=0 \).

Teacher's Note:
1. Divide the first by 3.
2. Subtract and multiply by 3/2.

Answer: (a) Isosceles

Explanation:
1. A point on the radical axis has equal powers, so \( PQ=PR \).
2. Two equal sides make the triangle isosceles.

Teacher's Note:
1. Points on the radical axis have equal tangent lengths.
2. Nothing forces the third side to be equal.

Answer: (d) -23

Explanation:
1. The common chord \( 4x+4y+k+15=0 \) must pass through the centre \( (-1,3) \) of the second circle.
2. \( -4+12+k+15=0 \), so \( k=-23 \).

Teacher's Note:
1. Bisecting the circumference means the common chord is a diameter.
2. Put the second centre in the radical axis.

Answer: (b) \( 2gx-2fy+g^2-f^2=0 \)

Explanation:
1. With equal radius \( r \): \( x^2+y^2+2gx+g^2-r^2=0 \) and \( x^2+y^2+2fy+f^2-r^2=0 \).
2. Subtracting: \( 2gx-2fy+g^2-f^2=0 \).

Teacher's Note:
1. Equal radii cancel on subtraction.
2. Subtract the two equations.