JEE Mathematics Circle and Conic Section MCQs Set 24

Answer: (c) \(15\)

Explanation:
1. Area \(=\tfrac12\cdot8\cdot y=4y\) with \(0<y<4\), so the area lies in \((0,16)\).
2. Integer areas \(1,2,\dots,15\) each give exactly one point in the first quadrant: 15 points.

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) \(4+2\sqrt2\)

Explanation:
1. The radius to \((2,-2)\) is perpendicular to \(x+y=0\), so it has slope 1; it meets the x-axis at the centre \((4,0)\).
2. Radius \(=\sqrt{4+4}=2\sqrt2\).
3. Greatest \(\alpha=4+2\sqrt2\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \((5\sqrt2-4,\infty)\)

Explanation:
1. Both curves are symmetric about the y-axis, so take \(x\ge0\): \(y=x+c\) and the circle \((x-4)^2+y^2=25\).
2. Tangency of \(y=x+c\) with the circle: \(\dfrac{|4+c|}{\sqrt2}=5\), giving \(c=5\sqrt2-4\) (the touching point has \(x>0\)).
3. There is no intersection for \(c>5\sqrt2-4\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (b) \(4x+3y+11=0\)

Explanation:
1. The tangent at \((0,-1)\) is \(y=-1\); the ray strikes it at \((-2,-1)\).
2. The reflected ray is a tangent to the circle through \((-2,-1)\) other than \(y=-1\): \(\dfrac{|2m-1|}{\sqrt{1+m^2}}=1\) gives \(m=\dfrac43\).
3. The incident ray has slope \(-\dfrac43\): \(y+1=-\dfrac43(x+2)\), i.e. \(4x+3y+11=0\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) \(3m-2=0\)

Explanation:
1. Substitute \(x=\dfrac{y-1}m\): \((1+m^2)y^2+(3m-2)y+(1-3m)=0\).
2. The two ordinates are equal and opposite, so their sum is zero: \(3m-2=0\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

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

Explanation:
1. Let the circle be \(x^2+y^2+2gx+2fy+c=0\). Orthogonality: \(2g\cdot0+2f\cdot0=c-4\), so \(c=4\).
2. Through \((a,b)\): \(a^2+b^2+2ga+2fb+4=0\).
3. Replace \((-g,-f)\) by \((x,y)\): \(2ax+2by=a^2+b^2+4\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (c) \(\dfrac9{\sqrt2}\)

Explanation:
1. The centre lies on the normal \(x+y=2\): centre \((1+t,1-t)\), radius \(\sqrt2|t|\), and the circle is \(x^2+y^2-2(1+t)x-2(1-t)y+2=0\).
2. Bisecting the circumference means the common chord passes through the centre \((0,-1)\) of the given circle. Common chord: \(-2(1+t)x+(2t-4)y+5=0\).
3. At \((0,-1)\): \(4-2t+5=0\), \(t=\dfrac92\), so the radius is \(\dfrac{9}{2}\sqrt2=\dfrac9{\sqrt2}\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) \(L_1<L_2\)

Explanation:
1. Substituting: \(x^2+8x-9=0\), so \(x=1\) and \(y=\pm2\sqrt2\).
2. \(L_1=4\sqrt2\approx5.66\) and \(L_2=8\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \((x_1,y_1)\)

Explanation:
1. Tangent: \(yy_1=2a(x+x_1)\), so \(x=\dfrac{yy_1}{2a}-x_1\).
2. Substitute: \(y^2-2y_1y+4ax_1-4ab=0\); the sum of ordinates is \(2y_1\), so the mid-point has \(y=y_1\).
3. Then \(x=\dfrac{y_1^2}{2a}-x_1=2x_1-x_1=x_1\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (c) \((x-1)^2+y^2=\dfrac94\)

Explanation:
1. Focus \((1,0)\), vertex \((0,0)\); the chord is \(x=\dfrac12\).
2. On the parabola \(y^2=2\); radius squared \(=\left(\dfrac12\right)^2+2=\dfrac94\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \(\dfrac{(x-2)^2}9+\dfrac{(y-1)^2}{25}=1\)

Explanation:
1. Centre \((2,1)\), \(2a=10\), \(2ae=8\), so \(a=5,\ ae=4,\ b^2=25-16=9\).
2. The major axis is vertical: \(\dfrac{(x-2)^2}9+\dfrac{(y-1)^2}{25}=1\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (b) \(4\)

Explanation:
1. \(c^2=16-9=7\), foci \((\pm\sqrt7,0)\).
2. \(r^2=7+9=16\), so \(r=4\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (c) \(3x^2+3y^2-6x-8=0\)

Explanation:
1. Squared distance of a point of the ellipse from \((1,0)\): \((x-1)^2+\dfrac{16-x^2}4=\dfrac34x^2-2x+5\).
2. Minimum at \(x=\dfrac43\): \(r^2=\dfrac{11}3\).
3. Circle: \((x-1)^2+y^2=\dfrac{11}3\), i.e. \(3x^2+3y^2-6x-8=0\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) \(y^2=-104x\)

Explanation:
1. Normal: \(\dfrac{16x}2-\dfrac{3y}{3/2}=13\), i.e. \(y=4x-\dfrac{13}2\).
2. The line \(y=mx+c\) touches \(y^2=4Ax\) if \(c=\dfrac Am\): \(-\dfrac{13}2=\dfrac A4\), \(A=-26\).
3. So the parabola is \(y^2=-104x\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \(\dfrac\pi2\)

Explanation:
1. Both ellipses have semi-axis 2 along the y-axis, so \(y=2\) touches both, at \((4,2)\) and \((-1,2)\).
2. Vectors from the origin: \((4,2)\cdot(-1,2)=-4+4=0\), so the angle is \(\dfrac\pi2\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (b) 4

Explanation:
1. For \(Y^2=8X\) the tangent is \(Y=mX+\dfrac2m\); with \(X=x+2\): \(c=2m+\dfrac2m\).
2. By AM-GM \(c\ge4\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) 3

Explanation:
1. Tangent \(y=mx+\dfrac1m\) through \((-2,-1)\): \(2m^2-m-1=0\), so \(m=1,\ -\dfrac12\).
2. \(|\tan\alpha|=\left|\dfrac{1+\frac12}{1-\frac12}\right|=3\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) 27

Explanation:
1. \(e=\dfrac23,\ ae=2\); ends of a latus rectum \(\left(2,\pm\dfrac53\right)\).
2. Tangent at \(\left(2,\dfrac53\right)\): \(\dfrac{2x}9+\dfrac y3=1\), with intercepts \(\dfrac92\) and 3.
3. The quadrilateral is a rhombus with half-diagonals \(\dfrac92\) and 3: area \(2\cdot\dfrac92\cdot3=27\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (c) 2

Explanation:
1. \(a=2,\ b=3,\ c^2=13\). Both pairs of foci are at distance \(\sqrt{13}\) from the centre, on perpendicular axes.
2. The quadrilateral is a square of diagonal \(2\sqrt{13}\): \(A=\dfrac12(2\sqrt{13})^2=26\), so \(\dfrac A{13}=2\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) 5

Explanation:
1. The tangent at \((2\sec\theta,\tan\theta)\) meets the asymptotes at \((2(\sec\theta+\tan\theta),\ \sec\theta+\tan\theta)\) and \((2(\sec\theta-\tan\theta),\ -(\sec\theta-\tan\theta))\).
2. \(CQ\cdot CR=\sqrt5\,|\sec\theta+\tan\theta|\cdot\sqrt5\,|\sec\theta-\tan\theta|=5\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \(c>0\)

Explanation:
1. Both circles pass through the points where \(x=y\) meets \(x^2+y^2+c=0\), i.e. \(2x^2+c=0\).
2. If one circle lies within the other they have no common point, so \(2x^2+c=0\) has no real root: \(c>0\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (c) \(\dfrac{25}3\)

Explanation:
1. Equal positive intercepts mean slope \(-1\). The normal at \((2\sec\theta,\tan\theta)\) is \(2x\cos\theta+y\cot\theta=5\); slope \(-1\) gives \(\sin\theta=\dfrac12\).
2. The normal becomes \(x+y=\dfrac5{\sqrt3}\), i.e. \(y=-x+\dfrac5{\sqrt3}\).
3. Tangency to the ellipse: \(c^2=a^2m^2+b^2\), so \(a^2+b^2=\dfrac{25}3\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \(\left(\dfrac\pi4,\pi\right]\)

Explanation:
1. The circle needs \(\sin\alpha\ge0\). The x-intercept is \(2\sqrt{g^2-c}=2\sqrt{\sin\alpha-\cos\alpha+1}\).
2. \(>2\) means \(\sin\alpha>\cos\alpha\), i.e. \(\alpha\in\left(\dfrac\pi4,\dfrac{5\pi}4\right)\).
3. Together with \(\sin\alpha\ge0\): \(\left(\dfrac\pi4,\pi\right]\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (b) \(K\in(0,2)\)

Explanation:
1. The fixed points are \((0,\pm1)\), at distance 2 apart.
2. For a hyperbola the difference of focal distances must be positive and less than the distance between the foci: \(0<K<2\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (c) 2

Explanation:
1. The angle at \((t,t)\) is a right angle, so the circle is \((x-1)(x-t)+(y-t)(y-1)=0\), i.e. \((x^2+y^2-x-y)+t(2-x-y)=0\).
2. For all \(t\): \(x+y=2\) and \(x^2+y^2=x+y=2\), giving \((1,1)\).
3. \(a^2+b^2=2\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (d) 4

Explanation:
1. \(PA\cdot PB=|t^2-4\cdot1|=|t^2-4|\) (power of the point).
2. \(|t^2-4|=3|t|\) gives \(|t|=4\) or \(|t|=1\).
3. Maximum \(t=4\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) 2

Explanation:
1. A chord exists only if \(\dfrac n{\sqrt2}<2\), i.e. \(n=1,2\).
2. Chord length squared \(=4\left(4-\dfrac{n^2}2\right)=16-2n^2\): 14 and 8.
3. Sum \(=22=11\cdot2\), so \(k=2\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (b) \(5x^2+5y^2-8x-16y-36=0\)

Explanation:
1. The radical axis of the two circles is \(4x+2y-5=0\); it meets \(12x-6y-41=0\) at \(\left(\dfrac73,-\dfrac{13}6\right)\).
2. For concyclic points the line \(AB\) also passes through this radical centre: \(\dfrac73-\dfrac{13}3+a=0\), \(a=2\).
3. The circle \(x^2+y^2-4-\dfrac85(x+2y+2)=0\) gives \(5x^2+5y^2-8x-16y-36=0\), which also equals \(S_2+\dfrac15(12x-6y-41)\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

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

Explanation:
1. \(a,b,c\) in AP means \(a-2b+c=0\): every line passes through \((1,-2)\). Chords bisected at their common point means it is the centre, so the centre is \((1,-2)\).
2. Auxiliary circle centre \((-\alpha,-\beta)\): \(\alpha=-1,\beta=2\); the circle is \(x^2+y^2-2x+4y+1=0\) with \(a^2=1+4-1=4\).
3. Semi-latus rectum \(\dfrac{b^2}a=\dfrac{2\cdot1\cdot3}{1+3}=\dfrac32\), so \(b^2=3\). Director circle radius squared \(a^2+b^2=7\): \(x^2+y^2-2x+4y-2=0\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (a) \(\dfrac12\)

Explanation:
1. Centre \((1,-2)\) and auxiliary circle radius \(a=2\).
2. Semi-latus rectum \(\dfrac{b^2}a=\dfrac32\) gives \(b^2=3\).
3. \(e^2=1-\dfrac{b^2}{a^2}=\dfrac14\), so \(e=\dfrac12\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.

Answer: (b) \((8,3,8)\)

Explanation:
1. The normal at \((t^2,2t)\) meets the x-axis at \(x=2+t^2\), so \(t=\pm1\): \(P(1,2),\ Q(1,-2)\); tangents meet at \(T(-1,0)\).
2. Diagonals \(TR=4\) and \(PQ=4\) are perpendicular: \(A=\dfrac12\cdot4\cdot4=8\).
3. Normals from \((3,0)\): \(t=0,\pm1\), so \(N=3\). \(PT^2=4+4=8\).

Teacher's Note:
1. Reduce the geometric condition to an algebraic one (distance, tangency, discriminant) before substituting.
2. For circles use centre and radius; for parabola, ellipse and hyperbola use standard parametric or tangent forms.