JEE Mathematics Circle MCQs Set 03

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

Explanation:
1. Let the midpoint be \( (h,k) \). The chord is \( (h-1)x+(k-1)y=h^2+k^2-h-k \).
2. Homogenising the circle with the chord, the \( xy \) coefficient must vanish for equally inclined lines, which gives \( (h-1)+(k-1)=0 \).
3. The locus is \( x+y=2 \).

Teacher's Note:
1. Equally inclined lines to the x-axis have slopes adding to zero.
2. Make the xy coefficient of the combined equation zero.

Answer: (d) none of these

Explanation:
1. Circle through the origin: \( x^2+y^2+2gx+2fy=0 \). Orthogonal: \( 2g\left(-\frac52\right)+2f\left(\frac32\right)=0+(-1) \Rightarrow-5g+3f=-1 \).
2. Centre \( (x,y)=(-g,-f) \): \( 5x-3y+1=0 \), a straight line of slope \( \frac53 \), not \( \frac35 \).
3. So none of the listed statements is correct.

Teacher's Note:
1. Orthogonality: 2gg' + 2ff' = c + c'.
2. Check the slope of the resulting line carefully.

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

Explanation:
1. \( C_1 \) has centre \( (2,1) \) and radius \( \sqrt{4+1+11}=4 \).
2. The centre of \( C_2 \) is always at distance \( 4+1=5 \) from \( (2,1) \).
3. \( (x-2)^2+(y-1)^2=25\Rightarrow x^2+y^2-4x-2y-20=0 \).

Teacher's Note:
1. External touching: distance between centres is the sum of radii.
2. Expand the locus circle.

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

Explanation:
1. The centre is equidistant from the two lines, so it lies on one of their angle bisectors.
2. \( \frac{x+2y}{\sqrt5}=\pm\frac{x-2y}{\sqrt5} \) gives \( y=0 \) or \( x=0 \). Together: \( xy=0 \).

Teacher's Note:
1. Equal distances from two lines give their bisectors.
2. Combine both bisectors into one equation.

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

Explanation:
1. Equal tangent lengths means the point lies on the radical axis.
2. Subtracting \( x^2+y^2-4=0 \) from \( x^2+y^2-5x+\frac32y-1=0 \): \( 5x-\frac32y-3=0 \), a straight line.
3. The radical axis is perpendicular to the line joining the centres.

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

Answer: (a) a line whose equation is \( x+2y=0 \)

Explanation:
1. The centres of circles through the two intersection points lie on the line joining the centres of the two circles.
2. Centres \( (0,0) \) and \( \left(1,-\frac12\right) \): the line is \( y=-\frac x2\Rightarrow x+2y=0 \).

Teacher's Note:
1. The centres lie on the perpendicular bisector of the common chord.
2. That line is the line of centres.

Answer: (b) \( (1,\sqrt3) \)

Explanation:
1. Consecutive vertices of a regular hexagon subtend \( 60^\circ \) at the centre.
2. The next vertex from \( (2,0) \) is \( (2\cos60^\circ,2\sin60^\circ)=(1,\sqrt3) \) (or \( (1,-\sqrt3) \) in the other direction).
3. Of the options only \( (1,\sqrt3) \) is a vertex.

Teacher's Note:
1. Each side subtends 60 degrees at the centre.
2. Use the polar form of the point.

Answer: (a) \( \frac{-1+\sqrt{29}}{14} \)

Explanation:
1. Equal chords: \( 1-d_1^2=5-d_2^2\Rightarrow d_2^2-d_1^2=4 \), where \( d_1,d_2 \) are distances of the centres \( (0,0) \) and \( (4,0) \) from the line \( mx-y+2m+\frac12=0 \).
2. \( \frac{(6m+\frac12)^2-(2m+\frac12)^2}{1+m^2}=4\Rightarrow\frac{32m^2+4m}{1+m^2}=4\Rightarrow7m^2+m-1=0 \).
3. \( m=\frac{-1\pm\sqrt{29}}{14} \). Option (a) is one of these (the other root is not listed).

Teacher's Note:
1. Equal chord lengths relate the distances of the centres.
2. Solve the resulting quadratic in m.

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

Explanation:
1. Centre \( \left(\frac12,-\frac32\right) \), \( r^2=\frac52 \). On \( L_2 \) the distance from the centre is \( \sqrt2 \), so the chord is \( 2\sqrt{\frac52-2}=\sqrt2 \) and the distance from the centre to \( L_1 \) is also \( \sqrt2 \).
2. For \( y=mx \): \( \frac{(m+3)^2}{4(1+m^2)}=2\Rightarrow7m^2-6m-1=0\Rightarrow m=1 \) or \( -\frac17 \).
3. So \( L_1 \) can be \( x-y=0 \) (the other line is \( x+7y=0 \), not listed).

Teacher's Note:
1. Equal chords mean equal distances from the centre.
2. Solve for the slope of the line through the origin.

Answer: (b) centre \( (2,0) \) and radius 3

Explanation:
1. \( (x-2)^2+y^2=9\cos^2\phi+9\sin^2\phi=9 \).
2. Centre \( (2,0) \), radius 3.

Teacher's Note:
1. Eliminate phi using sin squared + cos squared = 1.
2. Read the centre and radius from the standard form.

Answer: (a) \( (2-2\sqrt2,-3-2\sqrt2) \)

Explanation:
1. Centre \( (2,-3) \), \( r=4 \). The direction from the point to the centre is \( \frac{(-5,-5)}{5\sqrt2}=\left(-\frac1{\sqrt2},-\frac1{\sqrt2}\right) \).
2. The farthest point is the centre plus 4 units along this direction: \( (2-2\sqrt2,-3-2\sqrt2) \).

Teacher's Note:
1. The farthest point lies on the line through the external point and the centre, on the far side.
2. Move r units from the centre away from the point.

Answer: (a) polar line of the point \( (h,k) \) with respect to the circle

Explanation:
1. The polar of \( (h,k) \) is \( hx+ky=a^2 \) for any position of the point.
2. Since \( (h,k) \) is inside the circle, no real tangents exist, so it is not a real chord of contact nor a tangent.

Teacher's Note:
1. The polar exists for every point.
2. A chord of contact needs the point to be outside.

Answer: (b) the y-axis

Explanation:
1. For two members with parameters \( \lambda_1,\lambda_2 \), subtract the equations: \( 2(\lambda_1-\lambda_2)x=0\Rightarrow x=0 \).
2. The radical axis is the y-axis.

Teacher's Note:
1. Subtract the equations of the two circles.
2. The coefficients differ only in the x term.

Answer: (c) \( 16\pi \)

Explanation:
1. Perpendicular tangents to a circle meet on its director circle: \( C_2:\ x^2+y^2=8 \).
2. The same rule applied to \( C_2 \) gives \( C_3:\ x^2+y^2=16 \).
3. Area \( =\pi\cdot16=16\pi \).

Teacher's Note:
1. The director circle of x squared + y squared = r squared has radius r root 2.
2. Apply the rule twice.

Answer: (b) (1, 2)

Explanation:
1. Centre \( (2,-1) \), \( r^2=10 \). Parallel tangents: \( x-3y+c=0 \) with \( \frac{|2+3+c|}{\sqrt{10}}=\sqrt{10}\Rightarrow c=5 \) or \( -15 \).
2. For \( c=5 \) the contact point is the foot of the perpendicular from the centre: \( (2,-1)-\frac{10}{10}(1,-3)=(1,2) \). (The other tangent touches at \( (3,-4) \), which is not listed.)

Teacher's Note:
1. Parallel tangents are at distance r from the centre.
2. The contact point is the foot of the perpendicular from the centre.

Answer: (a) \( A=\left(\frac12,\frac{\sqrt3}2\right),B=\left(\frac12,-\frac{\sqrt3}2\right) \)

Explanation:
1. The chord of contact is \( 2x=1\Rightarrow x=\frac12 \).
2. Then \( y^2=1-\frac14\Rightarrow y=\pm\frac{\sqrt3}2 \).
3. The points of contact are \( \left(\frac12,\pm\frac{\sqrt3}2\right) \).

Teacher's Note:
1. Write the chord of contact first.
2. Intersect it with the circle.

Answer: (a) \( (\sqrt2-1)a \)

Explanation:
1. The four circles have centres \( (\pm a,\pm a) \) and radius a. A circle about the origin touching all four has radius \( R \) with \( R+a=\sqrt2\,a \) (touching externally).
2. \( R=(\sqrt2-1)a \), the smallest such circle.

Teacher's Note:
1. The centres are the corners of a square of side 2a.
2. The distance from the origin to a centre is a root 2.

Answer: (a) l is the polar of P with respect to C

Explanation:
1. The polar of \( (5,10) \) with respect to \( x^2+y^2=25 \) is \( 5x+10y=25\Rightarrow x+2y=5 \), which is l.
2. The distance of l from the origin is \( \sqrt5\lt 5 \), so l is not a tangent. For a secant through P, PQ is the harmonic mean of PA and PB, not the arithmetic mean.

Teacher's Note:
1. The polar of a point is found by T = 0.
2. A secant meets the polar at a harmonic conjugate point.

Answer: (a) \( x-2y+5=0 \)

Explanation:
1. The polar of \( (1,1) \): \( x+y-2(x+1)+(y+1)-4=0 \).
2. Simplifying: \( -x+2y-5=0\Rightarrow x-2y+5=0 \).

Teacher's Note:
1. Use T = 0 for the polar of a point.
2. Replace x squared by x x1 and x by (x + x1)/2.

Answer: (b) (-1,1)

Explanation:
1. The region is the upper half disc \( x^2+y^2\lt 5,\ y>0 \).
2. Need \( a+1>0 \Rightarrow a>-1 \) and \( a^2+(a+1)^2\lt 5\Rightarrow a^2+a-2\lt 0\Rightarrow -2\lt a\lt 1 \).
3. Intersection: \( -1\lt a\lt 1 \).

Teacher's Note:
1. Describe the region by inequalities.
2. Take the intersection of all conditions.

Answer: (c) 5

Explanation:
1. On the y-axis, \( 5x-2y+6=0 \) gives \( Q=(0,3) \).
2. The length of the tangent from Q is \( \sqrt{S_1}=\sqrt{0+9+0+18-2}=\sqrt{25}=5 \).

Teacher's Note:
1. Length of tangent equals the square root of the circle expression at the point.
2. Find Q from the given line first.

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

Explanation:
1. The line already touches the first circle. Its distance from \( (a,0) \) must also be b: \( (ma-b\sqrt{1+m^2})^2=b^2(1+m^2) \).
2. This reduces to \( ma=2b\sqrt{1+m^2} \), so \( m^2a^2=4b^2(1+m^2) \) and \( m^2=\frac{4b^2}{a^2-4b^2} \).
3. Positive value: \( m=\frac{2b}{\sqrt{a^2-4b^2}} \).

Teacher's Note:
1. Impose the tangency condition for the second circle.
2. Square both sides carefully.

Answer: (b) \( (6+4\sqrt3) \) sq. unit

Explanation:
1. The centres of the three coins form an equilateral triangle of side 2, and each centre is at distance 1 from the sides of the big triangle.
2. Each vertex of the big triangle is at distance \( \frac{1}{\tan30^\circ}=\sqrt3 \) along the side from the foot of the perpendicular from the nearest centre. So the side is \( s=2+2\sqrt3 \).
3. Area \( =\frac{\sqrt3}{4}(2+2\sqrt3)^2=\frac{\sqrt3}{4}(16+8\sqrt3)=6+4\sqrt3 \) sq. unit.

Teacher's Note:
1. The incircle-type offset at each corner is r cot 30 degrees.
2. Add the offsets at both ends to the distance between centres.

Answer: (a) 1

Explanation:
1. The circle is \( x^2+y^2-x-y=0 \).
2. \( \lambda^2+\frac1{\lambda^2}=\lambda+\frac1\lambda \). With \( t=\lambda+\frac1\lambda \): \( t^2-2=t\Rightarrow t=2 \) or \( -1 \).
3. \( t=-1 \) has no real \( \lambda \); \( t=2 \) gives \( \lambda=1 \).

Teacher's Note:
1. Find the circle through the three given points.
2. Use the symmetric substitution t = lambda + 1/lambda.

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

Explanation:
1. Putting \( y=x \): \( 2x^2-2x=0\Rightarrow x=0,1 \). So \( A=(0,0) \) and \( B=(1,1) \).
2. The circle on AB as diameter: \( x(x-1)+y(y-1)=0\Rightarrow x^2+y^2=x+y \).

Teacher's Note:
1. Find the end points of the intercept first.
2. Use the diameter form of a circle.

Answer: (b) \( 2ax+2by-(a^2+b^2+4)=0 \)

Explanation:
1. Let the circle be \( x^2+y^2+2gx+2fy+c=0 \). Orthogonality with \( x^2+y^2=4 \) gives \( c=4 \).
2. It passes through \( (a,b) \): \( a^2+b^2+2ga+2fb+4=0 \).
3. With centre \( (x,y)=(-g,-f) \): \( 2ax+2by-(a^2+b^2+4)=0 \).

Teacher's Note:
1. Use the orthogonality condition 2g g' + 2f f' = c + c'.
2. Replace g and f by the coordinates of the centre.