JEE Mathematics Circles MCQs Set 22

Answer: (b) \(x+4y+3=0\)

Explanation:
1. The centre: \(x-y=1,\ 2x+y=3\Rightarrow\left(\tfrac43,\tfrac13\right)\).
2. The radius to \((1,-1)\) has direction \(\left(-\tfrac13,-\tfrac43\right)\propto(1,4)\), so the tangent is \((x-1)+4(y+1)=0\Rightarrow x+4y+3=0\).

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Answer: (b) \(x^2+y^2-3x+9y=0\)

Explanation:
1. Putting \(y=-3x\): \(10x^2-30x=0\Rightarrow x=0,3\). The chord ends are \((0,0)\) and \((3,-9)\).
2. Circle on this diameter: \(x(x-3)+y(y+9)=0\Rightarrow x^2+y^2-3x+9y=0\).

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Answer: (b) \(\left(-\tfrac8{17},\tfrac2{17}\right)\)

Explanation:
1. The slope of the tangent at \(x=2\) is \(2x=4\): \(y-10=4(x-2)\Rightarrow4x-y+2=0\).
2. The circle's centre is \((-4,1)\). The point of contact is the foot of the perpendicular from the centre: \(4(-4)-1+2=-15\), so the foot is \(\left(-4,1\right)+\dfrac{15}{17}(4,-1)=\left(-\tfrac8{17},\tfrac2{17}\right)\).

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Answer: (d) \((25,29)\)

Explanation:
1. Inside: \(1+16-6-40+p \lt 0\Rightarrow p \lt 29\).
2. Centre \((3,5)\), \(r^2=34-p\). Not touching the axes requires \(r \lt 3\) (distance to the \(y\)-axis) \(\Rightarrow34-p \lt 9\Rightarrow p>25\); also \(r \lt 5\), automatically true.
3. So \(25 \lt p \lt 29\).

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Answer: (b) \((18,42)\)

Explanation:
1. Centres \((2,2)\), \((5,5)\); \(r_1=\sqrt2\), \(r_2=\sqrt{50-\lambda}\); distance \(3\sqrt2\).
2. Exactly two common tangents means the circles intersect: \(|r_2-\sqrt2| \lt 3\sqrt2 \lt r_2+\sqrt2\Rightarrow2\sqrt2 \lt r_2 \lt 4\sqrt2\).
3. \(8 \lt 50-\lambda \lt 32\Rightarrow18 \lt \lambda \lt 42\).

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Answer: (b) no common tangent

Explanation:
1. Centres \((0,0)\) and \((0,1)\); radii \(4\) and \(1\). The distance between centres \(=1 \lt 4-1=3\).
2. The second circle lies entirely inside the first, so there is no common tangent.

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Answer: (a) \(x^2+y^2+3x+y-11=0\)

Explanation:
1. Circles through the chord's end points: \(x^2+y^2-16+\lambda(3x+y+5)=0\).
2. The centre \(\left(-\tfrac{3\lambda}2,-\tfrac\lambda2\right)\) must lie on the chord: \(-\tfrac{9\lambda}2-\tfrac\lambda2+5=0\Rightarrow\lambda=1\).
3. Circle: \(x^2+y^2+3x+y-11=0\).

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Answer: (d) \(a=5\) and \(b\in(-\infty,1)\)

Explanation:
1. The centre is the intersection of the diameters: \(5x+8y=13,\ 4x-y=3\Rightarrow(1,1)\).
2. \(a^2-7a+11=1\Rightarrow a=2\) or \(5\); \(a^2-6a+6=1\Rightarrow a=1\) or \(5\). Hence \(a=5\).
3. The radius squared is \(1+1-(b^3+1)=1-b^3>0\Rightarrow b \lt 1\).

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Answer: (a) \(3x^2+3y^2-2\sqrt3ay=3a^2\)

Explanation:
1. \(C=(0,\sqrt3a)\). The circumcentre is the centroid \(\left(0,\tfrac{a}{\sqrt3}\right)\) and \(R=\dfrac{2a}{\sqrt3}\).
2. \(x^2+\left(y-\tfrac a{\sqrt3}\right)^2=\tfrac{4a^2}3\Rightarrow x^2+y^2-\tfrac{2a}{\sqrt3}y=a^2\Rightarrow3x^2+3y^2-2\sqrt3ay=3a^2\).

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Answer: (b) \(\pm4\)

Explanation:
1. Centre \((3,4)\), \(r^2=9+16-25+a^2=a^2\).
2. Touching the \(x\)-axis: \(r=4\Rightarrow a^2=16\Rightarrow a=\pm4\).

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Answer: (b) 6

Explanation:
1. The parallel tangents are \(\dfrac{16}{\sqrt2}=8\sqrt2\) apart, so \(r=4\sqrt2\). The centre lies midway: on \(y=x+2\).
2. Tangent to the \(y\)-axis in the first quadrant: \(p=r=4\sqrt2\), so \(q=4\sqrt2+2\).
3. \(p+q=2+8\sqrt2=a+b\sqrt a\) with \(a=2,\ b=8\): \(|a-b|=6\).

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Answer: (a) 1

Explanation:
1. Both circles pass through the origin, so they touch there: the centres \(\left(-\tfrac{3+\sin\theta}2,-\cos\phi\right)\) and \(\left(-\cos\phi,-\lambda\right)\) are collinear with the origin.
2. \(\dfrac{(3+\sin\theta)\lambda}2=\cos^2\phi\Rightarrow\lambda=\dfrac{2\cos^2\phi}{3+\sin\theta}\).
3. Maximum at \(\cos^2\phi=1,\ \sin\theta=-1\): \(\lambda=\dfrac22=1\).

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Answer: (b) 2

Explanation:
1. Orthogonality: \(2(g_1g_2+f_1f_2)=c_1+c_2\). Here \(g_1=1,f_1=\tfrac32,g_2=-\tfrac12,f_2=1\).
2. \(2\left(-\tfrac12+\tfrac32\right)=2=2c\Rightarrow c=1\), so \(2c=2\).

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Answer: (c) 15

Explanation:
1. \(x^2+2xy+3x+6y=(x+3)(x+2y)\): the normals are \(x=-3\) and \(x+2y=0\), meeting at the centre \(\left(-3,\tfrac32\right)\).
2. The circle to be contained has diameter from \((0,0)\) to \((4,3)\): centre \(\left(2,\tfrac32\right)\), radius \(\tfrac52\). Distance between centres \(=5\).
3. Just containing it: \(\lambda=5+\tfrac52=\tfrac{15}2\), so \(2\lambda=15\).

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Answer: (a) 1

Explanation:
1. The triangle is right angled at the origin, with vertices \((p,0),(0,q)\); its circumcentre is \(\left(\tfrac p2,\tfrac q2\right)=(x,y)\). The incircle has centre \((2,2)\) and radius \(2\).
2. The distance from \((2,2)\) to \(\dfrac xp+\dfrac yq=1\) equals \(2\), with \((2,2)\) inside: \(pq-2p-2q=2\sqrt{p^2+q^2}\).
3. With \(p=2x,q=2y\): \(xy-x-y=\sqrt{x^2+y^2}\), i.e. \(x+y-xy+\sqrt{x^2+y^2}=0\). So \(k=1\).

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Answer: (c) 35

Explanation:
1. Let \(\phi=x^2+y^2+2gx+2fy+c\). \(\phi(0,\lambda)=\lambda^2+2f\lambda+c\) has double root \(2\): \(f=-2,\ c=4\).
2. \(\phi(\lambda,0)=\lambda^2+2g\lambda+c\) has roots \(\tfrac45,5\) (product \(4=c\) ✓.) and sum \(-2g=\tfrac{29}5\Rightarrow g=-\tfrac{29}{10}\).
3. Centre \((\alpha,\beta)=\left(\tfrac{29}{10},2\right)\): \(10\alpha+3\beta=29+6=35\).

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Answer: (b) 6

Explanation:
1. The family passes through the points where \(y=0\) and \(x^2-2x-8=0\), independent of \(\lambda\).
2. \(x=4\) or \(-2\): \(A(4,0),\ B(-2,0)\), distance \(6\).

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Answer: (c) 4

Explanation:
1. The director circle of a circle of radius \(r\) has radius \(r\sqrt2\). So the radii are \(r_0,\ \tfrac{r_0}{\sqrt2},\ \tfrac{r_0}2,\ldots\), a G.P. with ratio \(\tfrac1{\sqrt2}\).
2. Sum \(=\dfrac{r_0}{1-\frac1{\sqrt2}}=r_0(2+\sqrt2)=2\Rightarrow r_0=2-\sqrt2\), \(r_0^2=6-4\sqrt2\).
3. For the given circle \(r_0^2=4+2-c=6-c\), so \(c=4\sqrt2\) and \(\lambda=4\).

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Answer: (b) 36

Explanation:
1. With centre distance \(d\): internal tangent\(^2=d^2-(r_1+r_2)^2=25\); external tangent\(^2=d^2-(r_1-r_2)^2=169\).
2. Subtracting: \(4r_1r_2=169-25=144\Rightarrow r_1r_2=36\).

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Answer: (a) 0

Explanation:
1. Chords through \((2,3)\) have distances from the centre \((3,4)\) ranging from \(0\) (the diameter) to \(\sqrt2\). The minimum distance is \(0\): the chord passes through the centre.
2. It passes through \((3,4)\) and \((2,3)\): \(3a+4b=5\) and \(2a+3b=5\Rightarrow a=-5,\ b=5\).
3. \(a^{2021}+b^{2021}=(-5)^{2021}+5^{2021}=0\).

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Answer: (b) 1

Explanation:
1. Both circles have radius \(3\) and centres \((1,2),(5,-6)\). A circle cutting both orthogonally has its centre on the radical axis (the perpendicular bisector of the centres, since the radii are equal).
2. Its radius squared equals the power of the centre, which is smallest at the midpoint \((3,-2)\) of the centres.
3. So \(a+b=3-2=1\).

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