JEE Mathematics Circle MCQs Set 02

Answer: (c) 9

Explanation:
1. Second circle: centre \( (4,3) \), \( r=\sqrt{25-n^2} \). The distance between centres is 5, and the first radius is 3.
2. Exactly two common tangents means the circles intersect at two points: \( |r-3|\lt 5\lt r+3 \), i.e. \( r>2 \) (and \( r\lt 8 \)).
3. \( 25-n^2>4\Rightarrow n^2\lt 21\Rightarrow n=-4,\dots,4 \), which is 9 values.

Teacher's Note:
1. Two common tangents means the circles cut at two points.
2. Count the integers satisfying the radius condition.

Answer: (c) 3

Explanation:
1. Centres \( (-1,-4) \) and \( (2,5) \); radii \( \sqrt{40}=2\sqrt{10} \) and \( \sqrt{10} \).
2. Distance between centres \( =\sqrt{9+81}=3\sqrt{10} \), which equals the sum of the radii.
3. The circles touch externally, so there are 3 common tangents.

Teacher's Note:
1. Sum of the radii equal to the centre distance means external contact.
2. External contact gives three common tangents.

Answer: (a) \( a^{-2}+b^{-2}=c^{-1} \)

Explanation:
1. Centres \( (-a,0),(0,-b) \), radii \( \sqrt{a^2-c},\sqrt{b^2-c} \). Touching externally: \( a^2+b^2=(\sqrt{a^2-c}+\sqrt{b^2-c})^2 \).
2. This gives \( c=\sqrt{(a^2-c)(b^2-c)} \), so \( c(a^2+b^2)=a^2b^2 \).
3. Dividing by \( a^2b^2c \): \( \frac1{a^2}+\frac1{b^2}=\frac1c \).

Teacher's Note:
1. Square the touching condition and simplify.
2. Divide by a squared b squared c.

Answer: (b) 1

Explanation:
1. First circle: through the origin with intercepts 2 and 2: centre \( (1,1) \), radius \( \sqrt2 \).
2. Second circle: centre \( \left(\frac12,\frac12\right) \), radius \( \frac{\sqrt2}2 \).
3. Distance between centres \( =\frac{\sqrt2}2=\sqrt2-\frac{\sqrt2}2 \), the difference of the radii, so they touch internally: one common tangent.

Teacher's Note:
1. Find both circles first.
2. Internal contact gives exactly one common tangent.

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

Explanation:
1. Second circle: centre \( (2\lambda,0) \), \( r^2=4\lambda^2-9 \), so \( |\lambda|>\frac32 \).
2. Two common tangents: the circles intersect, \( |r-2|\lt 2|\lambda|\lt r+2 \). The condition \( 2|\lambda|-2\lt r \) squares to \( 4\lambda^2-8|\lambda|+4\lt 4\lambda^2-9\Rightarrow|\lambda|>\frac{13}8 \).
3. So \( \lambda>\frac{13}8 \) or \( \lambda\lt -\frac{13}8 \).

Teacher's Note:
1. Two common tangents means two intersection points.
2. Use the distance between centres and the radii.

Answer: (a) they cut each other

Explanation:
1. Centres \( (2,3) \) and \( (1,0) \); radii \( \sqrt{21} \) and \( 2 \).
2. Distance between centres \( =\sqrt{10}\approx3.16 \). Since \( \sqrt{21}-2\approx2.58\lt 3.16\lt \sqrt{21}+2 \), the circles intersect.

Teacher's Note:
1. Compare the centre distance with the sum and the difference of radii.
2. In between means two intersection points.

Answer: (b) they cut each other orthogonally

Explanation:
1. First circle: centre \( (0,13) \), \( r_1^2=169-25=144 \). Second: centre \( O \), \( r_2^2=25 \).
2. Distance \( d^2=169=144+25=r_1^2+r_2^2 \).
3. The circles cut orthogonally.

Teacher's Note:
1. Orthogonal circles satisfy d squared = r1 squared + r2 squared.
2. Compute both radii and the centre distance.

Answer: (b) 2

Explanation:
1. The circle is \( x^2+y^2-\frac32x+2y=0 \).
2. \( S_1 \) at \( (2,1) \): \( 4+1-3+2=4 \).
3. \( PA=\sqrt{S_1}=2 \).

Teacher's Note:
1. Length of a tangent is the square root of S1.
2. Use the circle in standard form.

Answer: (c) (4, 1)

Explanation:
1. The tangent at the origin is \( -\frac x2+\frac y2=0\Rightarrow y=x \).
2. The reflection of \( (1,4) \) in the line \( y=x \) is \( (4,1) \).

Teacher's Note:
1. Find the tangent at the origin using the T = 0 form.
2. Reflection in y = x swaps the coordinates.

Answer: (a) \( y=m(x-1)+3\sqrt{1+m^2}-2 \)

Explanation:
1. Centre \( (1,-2) \), radius \( \sqrt{1+4+4}=3 \).
2. The tangents are \( y+2=m(x-1)\pm3\sqrt{1+m^2} \).
3. The upper sign gives option (a): \( y=m(x-1)+3\sqrt{1+m^2}-2 \).

Teacher's Note:
1. Shift the standard tangent form to the circle's centre.
2. Distance from the centre to the tangent equals the radius.

Answer: (c) \( 4\sqrt3 \)

Explanation:
1. \( OP=4,\ OA=2 \), so \( PA=\sqrt{16-4}=2\sqrt3 \).
2. Area \( =2\times\frac12\times OA\times PA=2\cdot2\sqrt3=4\sqrt3 \).

Teacher's Note:
1. The tangent is perpendicular to the radius at the contact point.
2. The quadrilateral is two congruent right triangles.

Answer: (b) \( \sin^{-1}\frac45 \)

Explanation:
1. Centre \( (-2,-1) \), \( r=3 \). \( S_1=1+\frac14+4+1-4=\frac94 \), so the tangent length is \( \frac32 \).
2. Half the angle at P satisfies \( \tan\frac\theta2=\frac r{\text{length}}=2 \), so \( \sin\theta=\frac{2\cdot2}{1+4}=\frac45 \).
3. The lines make an acute angle \( \sin^{-1}\frac45 \) (the angle at P itself is its supplement).

Teacher's Note:
1. The angle between a pair of lines is taken as the acute angle.
2. Use tan(theta/2) = r / tangent length.

Answer: (a) \( \left(\frac12,\frac14\right) \)

Explanation:
1. For a point \( (h,4-2h) \) the chord of contact is \( hx+(4-2h)y=1 \), i.e. \( h(x-2y)+(4y-1)=0 \).
2. It passes through the fixed point where \( x-2y=0 \) and \( 4y-1=0 \): \( \left(\frac12,\frac14\right) \).

Teacher's Note:
1. Write the chord of contact as a family in the parameter h.
2. Solve the two coefficient equations.

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

Explanation:
1. The normal through \( (4,3) \) and \( (2,1) \) is \( y=x-1 \) and passes through the centre.
2. Centre: \( 2x-y=2 \) and \( y=x-1 \) give \( (1,0) \).
3. Radius \( =\) distance to \( (2,1)=\sqrt2 \). Circle: \( (x-1)^2+y^2=2\Rightarrow x^2+y^2-2x-1=0 \).

Teacher's Note:
1. The normal to a circle passes through its centre.
2. Intersect the normal with the given diameter.

Answer: (b) 2

Explanation:
1. Every normal of a circle passes through its centre, so only the line joining the point to the centre can be a normal from the point.
2. That line meets the circle at two points, so there are 2 feet.

Teacher's Note:
1. Normals to a circle are radial lines.
2. A line through the centre cuts the circle at two points.

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

Explanation:
1. The centre is \( (2,0) \). The chord is perpendicular to the line from the centre to \( (1,1) \), which has slope \( -1 \), so the chord has slope 1.
2. \( y-1=x-1\Rightarrow x-y=0 \).

Teacher's Note:
1. A chord bisected at a point is perpendicular to the radius to that point.
2. Use T = S1 as a check.

Answer: (b) \( 4\sqrt3 \)

Explanation:
1. Coincident points means the line is a tangent, so PA is the length of the tangent from P.
2. \( S_1=4+9+4+30+1=48 \), so \( PA=\sqrt{48}=4\sqrt3 \).

Teacher's Note:
1. Tangent length is the square root of S1.
2. Substitute the point in the circle expression.

Answer: (c) 3

Explanation:
1. Let \( P=(t,t) \). Then \( PA^2=2t^2+2\lambda t+5 \) and \( PB^2=2t^2+2\lambda t+5 \).
2. They are equal, so \( PB=PA=3 \).

Teacher's Note:
1. Write both tangent lengths for the same point.
2. On the line x = y the two expressions coincide.

Answer: (a) \( \left(\frac85\sqrt6,\frac45\right) \)

Explanation:
1. A tangent through \( (0,5) \): \( y=mx+5 \) with \( \frac5{\sqrt{1+m^2}}=1\Rightarrow m=\pm2\sqrt6 \). Take \( m=-2\sqrt6 \): AB is \( 2\sqrt6x+y-5=0 \).
2. C is the pole of AB with respect to \( x^2+y^2=4 \): \( \left(-\frac{4l}{n},-\frac{4m'}{n}\right)=\left(\frac{8\sqrt6}5,\frac45\right) \).

Teacher's Note:
1. C is the pole of the chord AB.
2. Use the pole formula for the line lx + my + n = 0.

Answer: (c) \( \pm4\sqrt2 \)

Explanation:
1. Subtracting: \( -2\lambda y+\lambda^2=0\Rightarrow y=\frac\lambda2 \). The end points are \( \left(\pm\sqrt{16-\frac{\lambda^2}4},\frac\lambda2\right) \).
2. A right angle at O needs \( x_0^2=y_0^2 \): \( 16-\frac{\lambda^2}{4}=\frac{\lambda^2}{4}\Rightarrow\lambda^2=32 \).
3. \( \lambda=\pm4\sqrt2 \).

Teacher's Note:
1. Find the common chord by subtracting the equations.
2. Perpendicular vectors from the origin have zero dot product.

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

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

Teacher's Note:
1. The smallest circle uses the chord as a diameter.
2. The centre then lies on the line.

Answer: (a) \( \left(\frac7{2\sqrt2},\frac7{2\sqrt2}\right) \)

Explanation:
1. The line is at distance 5 from the origin. The nearest point of the circle to the line is \( (\sqrt2,\sqrt2) \), and the foot on the line is \( \left(\frac{5\sqrt2}2,\frac{5\sqrt2}2\right) \).
2. The smallest circle has these two points as ends of a diameter, so its centre is their midpoint: \( \frac{\sqrt2+\frac{5\sqrt2}2}2=\frac{7\sqrt2}4=\frac7{2\sqrt2} \).

Teacher's Note:
1. The smallest touching circle has its diameter along the common normal.
2. Take the midpoint of the two points.

Answer: (b) \( x^2+y^2=2x+4y \)

Explanation:
1. Substituting \( y=2x \): \( 5x^2-10x=0\Rightarrow x=0,2 \). The chord has ends \( (0,0) \) and \( (2,4) \).
2. Circle on this diameter: \( x(x-2)+y(y-4)=0\Rightarrow x^2+y^2=2x+4y \).

Teacher's Note:
1. Find the ends of the chord first.
2. Use the diameter form of a circle.

Answer: (a) \( x^2+y^2-4(x+y)-8\pi x+(2+4\pi)^2=0 \)

Explanation:
1. Initially the centre is \( (2,2) \). One complete rotation moves the centre \( 2\pi r=4\pi \) along the x-axis: new centre \( (2+4\pi,2) \).
2. \( (x-2-4\pi)^2+(y-2)^2=4\) expands to \( x^2+y^2-4(1+2\pi)x-4y+(2+4\pi)^2=0 \).
3. That is \( x^2+y^2-4(x+y)-8\pi x+(2+4\pi)^2=0 \).

Teacher's Note:
1. A rolling circle advances one circumference per turn.
2. Shift the centre and expand.

Answer: (c) \( 2x+2y=5 \)

Explanation:
1. Let the other end be \( (a,b) \) with \( a+b=3 \). The centre is \( (h,k)=\left(\frac{1+a}2,\frac{1+b}2\right) \).
2. \( 2h-1+2k-1=3\Rightarrow2h+2k=5 \).

Teacher's Note:
1. The centre is the midpoint of the diameter.
2. Eliminate the end point using its line.

Answer: (d) \( x^2+y^2+4x-6y+9=0 \)

Explanation:
1. Centre \( (-2,3) \); \( r^2=13-9\sin^2\alpha-13\cos^2\alpha=4\sin^2\alpha \), so \( r=2\sin\alpha \).
2. The half angle at P is \( \alpha \), so \( \sin\alpha=\frac rd\Rightarrow d=2 \).
3. P is at distance 2 from the centre: \( (x+2)^2+(y-3)^2=4\Rightarrow x^2+y^2+4x-6y+9=0 \).

Teacher's Note:
1. The half angle at P gives sin(alpha) = r / d.
2. The locus is a concentric circle.

Answer: (c) \( \text{a circle of radius }2\sqrt3 \)

Explanation:
1. For a chord of length 4, the distance of its midpoint from the centre is \( \sqrt{16-4}=2\sqrt3 \).
2. So the midpoint lies on the circle \( x^2+y^2=12 \), radius \( 2\sqrt3 \).

Teacher's Note:
1. The perpendicular from the centre bisects a chord.
2. Use Pythagoras with half the chord length.