JEE Mathematics Circles MCQs Set 27

Answer: (d) \( 210 \)

Explanation:
1. The distance of the line from the centre is \( \dfrac n{\sqrt2}\lt4 \), so \( n=1,2,3,4,5 \).
2. Chord length squared \( =4\left(16-\dfrac{n^2}2\right)=64-2n^2 \), and the sum over \( n=1..5 \) is \( 320-2(55)=210 \). Answer: (d)

Teacher's Note:
1. Chord length is \( 2\sqrt{r^2-d^2} \).

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

Explanation:
1. The tangent \( x\cos\theta+y\sin\theta=1 \) meets the axes at \( P(\sec\theta,0) \) and \( Q(0,\mathrm{cosec}\,\theta) \), so the mid-point is \( (h,k)=\left(\dfrac{\sec\theta}2,\dfrac{\mathrm{cosec}\,\theta}2\right) \).
2. \( \cos\theta=\dfrac1{2h} \), \( \sin\theta=\dfrac1{2k} \), and \( \sin^2\theta+\cos^2\theta=1 \) gives \( \dfrac1{4h^2}+\dfrac1{4k^2}=1 \), i.e. \( x^2+y^2=4x^2y^2 \). Answer: (c)

Teacher's Note:
1. Parametrise the tangent by the angle of its normal.

Answer: (a) \( (6,-2) \)

Explanation:
1. The centres are \( (0,0) \) and \( (-3,-4) \) with radii \( 2 \) and \( 7 \); the distance \( 5=7-2 \), so the circles touch internally and the common tangent is the radical axis.
2. Subtracting the equations gives \( 3x+4y=10 \), which is satisfied by \( (6,-2) \) only. Answer: (a)

Teacher's Note:
1. Internally touching circles have only one common tangent.

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

Explanation:
1. The circle is \( (x-1)^2+(y-1)^2+\lambda(x-y)=0 \). Passing through \( (1,-3) \): \( 16+4\lambda=0 \), so \( \lambda=-4 \).
2. The circle is \( x^2+y^2-6x+2y+2=0 \), with radius \( \sqrt{9+1-2}=2\sqrt2 \). Answer: (b)

Teacher's Note:
1. Use the family of circles touching a line at a given point.

Answer: (b) \( \dfrac{120}{13} \)

Explanation:
1. Orthogonal circles: the centres and a common point form a right triangle with legs \( 12 \) and \( 5 \), so \( C_1C_2=13 \).
2. The half-chord is the altitude to the hypotenuse: \( \dfrac{12\times5}{13}=\dfrac{60}{13} \), so the chord is \( \dfrac{120}{13} \). Answer: (b)

Teacher's Note:
1. At orthogonal intersection the radii to the common point are perpendicular.

Answer: (b) \( \dfrac1{\sqrt a}=\dfrac1{\sqrt b}+\dfrac1{\sqrt c} \)

Explanation:
1. For two externally tangent circles of radii \( r,s \) with a common tangent line, the distance between the points of contact is \( 2\sqrt{rs} \).
2. So \( 2\sqrt{ab}+2\sqrt{ac}=2\sqrt{bc} \) (the smallest circle lies between the other two). Dividing by \( 2\sqrt{abc} \): \( \dfrac1{\sqrt c}+\dfrac1{\sqrt b}=\dfrac1{\sqrt a} \). Answer: (b)

Teacher's Note:
1. Use the length of the common external tangent \( 2\sqrt{rs} \).

Answer: (a) \( 5 \)

Explanation:
1. The given circle has centre \( (-2,3) \) and radius \( 5 \). The centre of \( C \) lies on the line through \( (-2,3) \) and \( (1,-1) \), at \( (1,-1)+s\left(\dfrac35,-\dfrac45\right) \) with radius \( s \) (external contact).
2. Passing through \( (4,0) \): \( (3-0.6s)^2+(1+0.8s)^2=s^2 \), which gives \( 10-2s=0 \), so \( s=5 \). Answer: (a)

Teacher's Note:
1. At the point of contact the two centres and the point are collinear.

Answer: (c) \( \sqrt{41} \)

Explanation:
1. The circle has centre \( (3,-4) \) and radius \( 8\sqrt2 \), so a square with sides parallel to the axes has half-side \( 8 \) and vertices \( (3\pm8,-4\pm8) \).
2. The vertices are \( (11,4) \), \( (-5,4) \), \( (11,-12) \), \( (-5,-12) \), at distances \( \sqrt{137},\sqrt{41},\sqrt{265},13 \); the nearest is \( \sqrt{41} \). Answer: (c)

Teacher's Note:
1. The half-diagonal equals the radius, so the half-side is \( r/\sqrt2 \).

Answer: (a) a parabola

Explanation:
1. Let the circle be \( x^2+y^2+2gx+2fy+c=0 \). The chord condition \( 4a=2\sqrt{g^2-c} \) gives \( c=g^2-4a^2 \), and passing through \( (0,2b) \) gives \( 4b^2+4bf+c=0 \).
2. Eliminating \( c \): \( 4b^2+4bf+g^2-4a^2=0 \). With the centre \( (x,y)=(-g,-f) \): \( x^2=4by+4a^2-4b^2 \), a parabola. Answer: (a)

Teacher's Note:
1. The locus has one squared and one linear variable.

Answer: (c) \( [12,21] \)

Explanation:
1. The centres are \( (1,1) \) and \( (9,1) \) with radii \( 1 \) and \( 2 \). Opposite sides: \( (7-\lambda)(31-\lambda)\lt0 \), so \( 7\lt\lambda\lt31 \).
2. The line must not cut either circle: \( \dfrac{|7-\lambda|}5\ge1 \) gives \( \lambda\ge12 \) (in the range), and \( \dfrac{|31-\lambda|}5\ge2 \) gives \( \lambda\le21 \). Hence \( \lambda\in[12,21] \). Answer: (c)

Teacher's Note:
1. The line must keep each circle entirely on one side.

Answer: (c) \( 3\sqrt{\dfrac52} \)

Explanation:
1. The orthocentre, centroid and circumcentre are collinear with the centroid dividing \( HO \) in the ratio \( 2:1 \), so \( AB:BC=2:1 \). Here \( AB=\sqrt{36+4}=2\sqrt{10} \), so \( BC=\sqrt{10} \).
2. \( AC=3\sqrt{10} \), and the radius is \( \dfrac{3\sqrt{10}}2=3\sqrt{\dfrac52} \). Answer: (c)

Teacher's Note:
1. This is Euler's line property.

Answer: (d) a parabola

Explanation:
1. The given circle has centre \( (4,4) \) and radius \( 6 \). If the centre is \( (x,y) \), the radius is \( |y| \) and external contact gives \( \sqrt{(x-4)^2+(y-4)^2}=6+|y| \).
2. This gives \( (x-4)^2=20(y+1) \) for \( y\gt0 \) (and \( (x-4)^2=-4(y-5) \) for \( y\lt0 \)), both parabolas. Answer: (d)

Teacher's Note:
1. Distance from the fixed centre equals the sum of radii.

Answer: (c) 3

Explanation:
1. The centres are \( (2,3) \) and \( (-3,-9) \) with radii \( 5 \) and \( 8 \); the distance between the centres is \( \sqrt{25+144}=13=5+8 \).
2. The circles touch externally, so they have \( 3 \) common tangents. Answer: (c)

Teacher's Note:
1. Compare the distance between centres with the sum and difference of radii.

Answer: (c) \( (5,-2) \)

Explanation:
1. The circle is \( (x-3)^2+y^2+\lambda y=0 \). Passing through \( (1,-2) \): \( 4+4-2\lambda=0 \), so \( \lambda=4 \) and the circle is \( (x-3)^2+y^2+4y=0 \).
2. Checking the options, \( (5,-2) \) gives \( 4+4-8=0 \), while the others do not satisfy it. Answer: (c)

Teacher's Note:
1. Use the family of circles touching a line at a given point.

Answer: (a) \( |a|=c \)

Explanation:
1. The circles have centres \( \left(\dfrac a2,0\right) \) and \( (0,0) \) with radii \( \dfrac{|a|}2 \) and \( c \). External touching would need \( \dfrac{|a|}2+c=\dfrac{|a|}2 \), impossible for \( c\gt0 \).
2. So they touch internally: the distance between the centres equals the difference of the radii, \( \dfrac{|a|}2=c-\dfrac{|a|}2 \), so \( |a|=c \). Answer: (a)

Teacher's Note:
1. Since the first circle passes through the origin, only internal contact is possible.

Answer: (b) \( -35\lt m\lt15 \)

Explanation:
1. The circle has centre \( (2,4) \) and radius \( 5 \). Two distinct intersections need the distance from the centre to the line to be less than the radius.
2. \( \dfrac{|6-16-m|}5\lt5 \) gives \( |m+10|\lt25 \), so \( -35\lt m\lt15 \). Answer: (b)

Teacher's Note:
1. Compare the distance from the centre with the radius.

Answer: (b) \( k\ge\dfrac12 \)

Explanation:
1. A circle tangent to the \( x \)-axis with centre \( (h,k) \) has radius \( |k| \): passing through \( (-1,1) \) gives \( (h+1)^2+(1-k)^2=k^2 \), i.e. \( h^2+2h-2k+2=0 \).
2. For real \( h \) the discriminant is non-negative: \( 4-4(2-2k)\ge0 \), so \( k\ge\dfrac12 \). Answer: (b)

Teacher's Note:
1. Real \( h \) requires a non-negative discriminant.

Answer: (c) \( x^2+y^2=\dfrac94 \)

Explanation:
1. The mid-point \( P \) of a chord \( AB \) satisfies \( OP\perp AB \) and \( \angle AOP=\dfrac\pi3 \), so \( OP=OA\cos\dfrac\pi3=\dfrac32 \).
2. Hence \( x^2+y^2=\dfrac94 \). Answer: (c)

Teacher's Note:
1. The line from the centre to a chord's mid-point bisects the angle at the centre.

Answer: (a) a parabola

Explanation:
1. Let the centre be \( (h,k) \) with radius \( |k| \). External contact: \( h^2+(k-3)^2=(|k|+2)^2 \), so \( h^2=10k-5 \) for \( k\gt0 \).
2. This is a parabola. Answer: (a)

Teacher's Note:
1. The distance between the centres equals the sum of the radii.

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 gives \( c=4 \), and \( (a,b) \) on it gives \( a^2+b^2+2ga+2fb+4=0 \).
2. With the centre \( (-g,-f)=(x,y) \): \( 2ax+2by-(a^2+b^2+4)=0 \). Answer: (b)

Teacher's Note:
1. Orthogonal circles satisfy \( 2g_1g_2+2f_1f_2=c_1+c_2 \).

Answer: (a) \( x^2+y^2-2x+2y-23=0 \)

Explanation:
1. The diameters meet at the centre \( (1,-1) \), and \( 2\pi r=10\pi \) gives \( r=5 \).
2. The circle is \( (x-1)^2+(y+1)^2=25 \), i.e. \( x^2+y^2-2x+2y-23=0 \). Answer: (a)

Teacher's Note:
1. The centre lies on both diameters.

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

Explanation:
1. Substituting \( y=x \): \( 2x^2-2x=0 \), so \( A=(0,0) \) and \( B=(1,1) \).
2. The circle on \( AB \) as diameter is \( x(x-1)+y(y-1)=0 \), i.e. \( x^2+y^2-x-y=0 \). Answer: (a)

Teacher's Note:
1. Use the diameter form \( (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 \).

Answer: (a) \( 2\lt r\lt8 \)

Explanation:
1. The centres are \( (1,3) \) and \( (4,-1) \) at distance \( 5 \); the radii are \( r \) and \( \sqrt{17-8}=3 \).
2. Two intersection points need \( |r-3|\lt5\lt r+3 \), which gives \( 2\lt r\lt8 \). Answer: (a)

Teacher's Note:
1. Use \( |r_1-r_2|\lt C_1C_2\lt r_1+r_2 \).

Answer: (c) \( x^2+y^2-2x+2y=47 \)

Explanation:
1. The diameters meet at the centre \( (1,-1) \). Area \( \pi r^2=154 \) with \( \pi=\dfrac{22}7 \) gives \( r=7 \).
2. \( (x-1)^2+(y+1)^2=49 \), i.e. \( x^2+y^2-2x+2y=47 \). Answer: (c)

Teacher's Note:
1. Take \( \pi=22/7 \) as the problem intends.

Answer: (b) 15 units

Explanation:
1. The circle has centre \( C(2,1) \) and radius \( 5 \), and \( PC=\sqrt{64+36}=10 \).
2. The greatest distance from \( P \) to the circle is \( PC+r=10+5=15 \). Answer: (b)

Teacher's Note:
1. The farthest point on a circle lies on the line through the centre.

Answer: (b) \( x+y=2\sqrt2 \)

Explanation:
1. The circle has centre \( (-2,2) \) and radius \( 2 \). A line with equal positive intercepts is \( x+y=a \).
2. Tangency: \( \dfrac{|-2+2-a|}{\sqrt2}=2 \), so \( a=2\sqrt2 \). Answer: (b)

Teacher's Note:
1. The distance from the centre to a tangent equals the radius.