JEE Mathematics Circles MCQs Set 17

Answer: (d) A straight line perpendicular to the line joining the centres of the circles

Explanation:
1. Equal tangent lengths: the radical axis. The second circle is \( x^2+y^2-5x+\frac32y-1=0 \).
2. Subtracting: \( 5x-\frac32y-3=0 \), i.e. \( 10x-3y-6=0 \).
3. The radical axis is a straight line perpendicular to the line of centres.

Teacher's Note:
1. Equal tangent lengths mean equal powers.
2. The radical axis is always perpendicular to the line of centres.

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

Explanation:
1. Subtracting: \( -2ax-2by+2bx+2ay=0 \), i.e. \( (b-a)(x-y)=0 \).
2. For \( a\neq b \) the radical axis is \( x-y=0 \).

Teacher's Note:
1. The squared terms and c² cancel.
2. Factor the remaining terms.

Answer: (b) Passes through the centre of \( S_2 \)

Explanation:
1. The common chord is a diameter of \( S_2 \), so the radical axis passes through the centre of \( S_2 \).

Teacher's Note:
1. Bisecting the circumference means the chord is a diameter of S₂.
2. The radical axis is the line of the common chord.

Answer: (b) Passes through the centre of the third circle

Explanation:
1. If the third circle has centre \( C \) and radius \( \rho \), orthogonality means the power of \( C \) with respect to each of the two circles equals \( \rho^2 \).
2. So \( C \) has equal powers with respect to both circles, i.e. it lies on their radical axis.

Teacher's Note:
1. For orthogonal circles, the square of the radius equals the power of the centre.
2. Equal powers put the centre on the radical axis.

Answer: (b) 10

Explanation:
1. The radical axis is \( 2x-2y+p-14=0 \).
2. Through \( (1,-1) \): \( 2+2+p-14=0 \), so \( p=10 \).

Teacher's Note:
1. Subtract the two circles.
2. Substitute the point.

Answer: (a) \( 21x-68=0 \)

Explanation:
1. Divide: \( x^2+y^2-x+2y+\frac{18}7=0 \) and \( x^2+y^2-\frac74x+2y+5=0 \).
2. Subtracting: \( \frac34x-\frac{17}7=0 \), i.e. \( 21x-68=0 \).

Teacher's Note:
1. Make the coefficients of x² and y² equal to 1 first.
2. Subtract and clear fractions.

Answer: (b) \( d_1-d_3=5 \)

Explanation:
1. Equal powers at \( (1,2) \): \( -10+d_1=-4+d_2=-5+d_3 \).
2. So \( d_1-10=d_3-5 \), which gives \( d_1-d_3=5 \).

Teacher's Note:
1. The radical centre has equal power with respect to all three circles.
2. Substitute (1, 2) in each.

Answer: (b) \( \left(\frac3{25},\frac4{25}\right) \)

Explanation:
1. The radical axis of the circle and the point circle \( x^2+y^2=0 \) is \( -6x-8y+1=0 \). The system is \( x^2+y^2+\mu(-6x-8y+1)=0 \).
2. Centre \( (3\mu,4\mu) \) and \( r^2=25\mu^2-\mu \). Point circles: \( r=0 \), so \( \mu=0 \) (the origin) or \( \mu=\frac1{25} \).
3. The other limiting point is \( \left(\frac3{25},\frac4{25}\right) \).

Teacher's Note:
1. Limiting points are the point circles of the system.
2. Set r² = 0 in the family.

Answer: (a) (-1, 1)

Explanation:
1. The radical axis is \( 4x+2y-1=0 \). The family is \( S_1+\lambda(4x+2y-1)=0 \) with centre \( (3-2\lambda,3-\lambda) \).
2. Point circles: \( (3-2\lambda)^2+(3-\lambda)^2-4+\lambda=0 \), i.e. \( 5\lambda^2-17\lambda+14=0 \), so \( \lambda=2 \) or \( \frac75 \).
3. For \( \lambda=2 \) the centre is \( (-1,1) \).

Teacher's Note:
1. A limiting point is the centre of a zero-radius member.
2. Compute r² = g² + f² − c for the family.

Answer: (a) Intersecting circles

Explanation:
1. Putting \( x=0 \): \( y^2=-c>0 \), so every member passes through the two fixed points \( (0,\pm\sqrt{-c}) \).
2. The circles therefore intersect in two points.

Teacher's Note:
1. Look for fixed common points.
2. c < 0 gives real values of y.

Answer: (a) \( A,B,C,D \) are concyclic

Explanation:
1. The points are \( (b,0),(0,a) \) and \( (a,0),(0,b) \).
2. The product of the intercepts on the x-axis is \( ab \), the same as on the y-axis (\( a\cdot b \)), so the four points are concyclic.

Teacher's Note:
1. Use the power of the origin.
2. OA · OC = OB · OD.

Answer: (b) 0

Explanation:
1. For concyclic points \( P,Q,R,S \), the lines \( PQ \), \( RS \) and the radical axis of the two circles are concurrent.
2. The radical axis is \( (a-a')x+(b-b')y+(c-c')=0 \).
3. Concurrency of the three lines means \( D=0 \).

Teacher's Note:
1. Three chords/axes of a concyclic configuration meet at the radical centre.
2. Concurrent lines give a zero determinant.

Answer: (a) \( 180^\circ \)

Explanation:
1. The slopes are \( -\frac1{\sqrt3} \) and \( \sqrt3 \), whose product is \( -1 \): the lines are perpendicular.
2. The circle passes through their intersection, so the angle at that point on the circle is \( 90^\circ \), making \( PQ \) a diameter.
3. The arc \( PQ \) subtends \( 180^\circ \) at the centre.

Teacher's Note:
1. Check whether the lines are perpendicular.
2. A right angle in a semicircle means PQ is a diameter.

Answer: (b) A circle

Explanation:
1. The centroid is \( G=\frac{P+A+B}3 \), where \( A \) and \( B \) are fixed.
2. As \( P \) moves on a circle of radius \( r \), \( G \) moves on a circle of radius \( \frac r3 \) with centre \( \frac{A+B}3 \).

Teacher's Note:
1. Centroid = average of the three vertices.
2. A fixed shift and a scaling by 1/3 maps a circle to a circle.

Answer: (a) \( a_1a_2=b_1b_2 \)

Explanation:
1. The intercepts are \( -\frac{c_1}{a_1},-\frac{c_2}{a_2} \) on the x-axis and \( -\frac{c_1}{b_1},-\frac{c_2}{b_2} \) on the y-axis.
2. Concyclic: the products of intercepts are equal: \( \frac{c_1c_2}{a_1a_2}=\frac{c_1c_2}{b_1b_2} \), so \( a_1a_2=b_1b_2 \).

Teacher's Note:
1. Use the power of the origin.
2. Products of the intercepts on the two axes are equal.

Answer: (a) (3, 0)

Explanation:
1. \( Q \) is the reflection of \( P(p,q) \) in \( x-y+1=0 \), i.e. \( Q=(q-1,p+1) \).
2. \( Q \) on \( 7x+y+3=0 \): \( 7(q-1)+(p+1)+3=0 \), so \( p=3-7q \). With \( p^2+q^2=9 \): \( 50q^2-42q=0 \), so \( q=0 \) or \( q=\frac{21}{25} \).
3. The points are \( (3,0) \) and \( \left(-\frac{72}{25},\frac{21}{25}\right) \); positive abscissa gives \( (3,0) \).

Teacher's Note:
1. The perpendicular bisector is the mirror line of P and Q.
2. Use the reflection in x − y + 1 = 0: (x, y) → (y − 1, x + 1).

Answer: (c) 72

Explanation:
1. \( P \) lies on \( y=x \) with \( OP=6\sqrt2 \), so \( P=(6,6) \).
2. Substituting \( y=x \): \( 2x^2+2(g+f)x+c=0 \) has a double root \( x=6 \), so the product of roots \( \frac c2=36 \).
3. \( c=72 \).

Teacher's Note:
1. A tangent meets the circle at a double root.
2. Product of the roots = c/2.

Answer: (b) Two

Explanation:
1. Two rational points \( (x_1,y_1),(x_2,y_2) \) on the circle satisfy \( x_1^2+y_1^2-2\sqrt3x_1=x_2^2+y_2^2-2\sqrt3x_2 \).
2. Since \( \sqrt3 \) is irrational, \( x_1=x_2 \), and then \( y_1^2=y_2^2 \), so \( y_2=-y_1 \).
3. A third rational point would repeat one of these values, so the maximum is 2.

Teacher's Note:
1. Use the irrationality of √3.
2. Rational points can differ only in the sign of y.

Answer: (a) 7

Explanation:
1. On \( C_k \), an arc of length \( k \) subtends \( 1 \) radian. So after the motion on \( C_n \) the total angle turned is \( n \) radians.
2. The positive \( x\text{-axis} \) is crossed at \( 2\pi\approx6.28 \) radians. After 6 steps the angle is 6 (not yet), and during the 7th step it passes 6.28.
3. So \( n=7 \).

Teacher's Note:
1. Arc length = radius × angle.
2. Each step turns by exactly 1 radian.

Answer: (c) A circle of radius \( \sqrt{a^2-r^2} \)

Explanation:
1. For a regular polygon of \( n \) vertices with circumcentre \( O \): \( \sum PV_i^2=n(OP^2+r^2) \).
2. \( 6(OP^2+r^2)=6a^2 \), so \( OP^2=a^2-r^2 \).
3. The locus is a circle of radius \( \sqrt{a^2-r^2} \).

Teacher's Note:
1. Use the sum of squares of distances formula for a regular polygon.
2. The locus is centred at the circumcentre.

Answer: (b) \( \sqrt3y=x+4 \)

Explanation:
1. \( P \) is at \( 30^\circ \); after a quarter circle it is at \( 120^\circ \): \( (-1,\sqrt3) \).
2. The tangent there is \( -x+\sqrt3y=4 \), i.e. \( \sqrt3y=x+4 \).

Teacher's Note:
1. A quarter of the circle is 90°.
2. Tangent at (x₁, y₁): xx₁ + yy₁ = 4.

Answer: (b) -4

Explanation:
1. A conic through the four points is \( S_1+\lambda S_2=0 \). For a circle the \( xy \) term vanishes: \( 4+6\lambda=0 \), so \( \lambda=-\frac23 \).
2. Then the coefficients of \( x^2 \) and \( y^2 \) must be equal: \( a\left(1-\frac23\right)=2-\frac{10}3 \), i.e. \( \frac a3=-\frac43 \).
3. So \( a=-4 \).

Teacher's Note:
1. The family S₁ + λS₂ = 0 contains all conics through the four points.
2. A circle has no xy term and equal x², y² coefficients.