JEE Mathematics Circle and Conic Section MCQs Set 23

Answer: (b) \( \left(\frac45,\frac75\right) \)

Explanation:
1. The given circle has centre (2, 3) and radius \( \sqrt{4+9+12}=5 \). The point (−1, −1) is at distance 5 from (2, 3).
2. The required centre lies on the line from (2, 3) to (−1, −1) at distance 5 − 3 = 2 from (2, 3).
3. Direction \( \frac{(-3,-4)}5 \), so the centre is \( (2,3)+2\cdot\frac{(-3,-4)}5=\left(\frac45,\frac75\right) \).

Teacher's Note:
1. For internal contact the centres and the point of contact are collinear.
2. Distance between centres = difference of radii.

Answer: (c) \( 5x^2+5y^2+60x+7=0 \)

Explanation:
1. The square of the tangent length is S₁ and S₂ respectively, so \( \frac{S_1}{S_2}=\frac49 \).
2. Then \( 9(x^2+y^2+4x+3)=4(x^2+y^2-6x+5) \).
3. Simplifying: \( 5x^2+5y^2+60x+7=0 \).

Teacher's Note:
1. Length of tangent from (x, y) = √S₁.
2. Square the ratio.

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

Explanation:
1. For \( x\ge0 \) the circle is \( (x-4)^2+y^2=25 \) and the line is \( y=x+c \); for \( x<0 \) both are mirror images, so it is enough to study \( x\ge0 \).
2. The line y = x + c touches this circle when \( \frac{|4+c|}{\sqrt2}=5 \), i.e. \( c=5\sqrt2-4 \) (upper tangent) or \( c=-5\sqrt2-4 \).
3. For very negative c the line meets the lower part of the circle, and for \( c\le5\sqrt2-4 \) there is always an intersection with \( x\ge0 \); beyond the upper tangent, \( c>5\sqrt2-4 \), the line misses the circle. So there is no solution for \( c\in(5\sqrt2-4,\infty) \).

Teacher's Note:
1. Reduce to x ≥ 0 using symmetry.
2. Find the c at which the line y = x + c is tangent to the circle.

Answer: (c) Statement-I is true but Statement-II is false.

Explanation:
1. Let the chords of contact (polars of A, B, C) meet at R. By the pole-polar relation, A, B, C all lie on the polar of R, which is a line, so they are collinear: Statement-I is true.
2. Three arbitrary collinear points need not lie on a normal (a line through the centre): Statement-II is false.

Teacher's Note:
1. Polar of a point and the concurrence of polars.
2. A normal of a circle passes through its centre.

Answer: (d) Statement-I is false but Statement-II is true.

Explanation:
1. The centres are (1, 2) and (2, 1), at distance \( \sqrt2 \); the radii are 5 and 4. Since \( 5-4=1<\sqrt2<9 \), the circles intersect, neither lying inside the other.
2. A circle of radius r touching both has centre P with \( PC_1=5\pm r \) and \( PC_2=4\pm r \): the sum or the difference of the distances is constant, so the locus consists of an ellipse and a hyperbola. Statement-I is false.
3. If S₂ lies completely inside S₁: \( PC_1=R_1-r \) and \( PC_2=R_2+r \), so \( PC_1+PC_2=R_1+R_2 \) (more than the distance between the centres): an ellipse. Statement-II is true.

Teacher's Note:
1. Find the distance between the centres and compare it with the radii.
2. Sum of focal distances constant gives an ellipse, difference a hyperbola.

Answer: (c) (2, 1)

Explanation:
1. The nearest point is where the tangent is parallel to x + y = 7. The tangent at \( (x_1,y_1) \) is \( xx_1+2yy_1=6 \), parallel when \( x_1=2y_1 \).
2. Substituting in the ellipse: \( 4y_1^2+2y_1^2=6 \), so \( y_1=\pm1 \) and the points are (2, 1) and (−2, −1).
3. (2, 1) is on the same side as the line and is nearer (distance \( \frac{4}{\sqrt2} \) against \( \frac{10}{\sqrt2} \)).

Teacher's Note:
1. Parallel tangent gives the nearest and farthest points.
2. Compare the two candidate points.

Answer: (c) \( \frac{\sqrt3}2 \)

Explanation:
1. Latus rectum \( =\frac{2b^2}a \) and minor axis \( =2b \).
2. Given \( \frac{2b^2}a=\frac12\cdot2b \), so \( b=\frac a2 \).
3. Then \( e^2=1-\frac{b^2}{a^2}=\frac34 \) and \( e=\frac{\sqrt3}2 \).

Teacher's Note:
1. Latus rectum of an ellipse = 2b²/a.
2. e² = 1 − b²/a².

Answer: (a) \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{ex}a \)

Explanation:
1. Let the mid-point be (h, k). The chord with this mid-point is \( T=S_1 \): \( \frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2} \).
2. It passes through the focus (ae, 0): \( \frac{he}a=\frac{h^2}{a^2}+\frac{k^2}{b^2} \).
3. Replacing (h, k) by (x, y): \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{ex}a \).

Teacher's Note:
1. Use the chord with a given mid-point: T = S₁.
2. Substitute the focus into it.

Answer: (b) \( -\frac{5\pi}6 \)

Explanation:
1. The ends of a diameter are \( (a\cos\theta,b\sin\theta) \) and \( (-a\cos\theta,-b\sin\theta) \), i.e. their eccentric angles differ by \( \pi \).
2. So B has eccentric angle \( \frac\pi6+\pi=\frac{7\pi}6 \), which is equivalent to \( -\frac{5\pi}6 \).

Teacher's Note:
1. B is the reflection of A in the centre.
2. Eccentric angles of the ends of a diameter differ by π.

Answer: (d) 27 sq unit

Explanation:
1. Here \( a^2=9,\ b^2=5 \), \( e=\frac23 \), \( ae=2 \), and the latus rectum ends are \( \left(\pm2,\pm\frac53\right) \).
2. The tangent at \( \left(2,\frac53\right) \) is \( \frac{2x}9+\frac y3=1 \), cutting the axes at \( \left(\frac92,0\right) \) and (0, 3). The four tangents form a rhombus with diagonals 9 and 6.
3. Area \( =\frac12\cdot9\cdot6=27 \) sq unit.

Teacher's Note:
1. Write the tangent at (ae, b²/a).
2. The quadrilateral is a rhombus; use its diagonals.

Answer: (a) ellipse

Explanation:
1. The roots are non-real when \( b^2-ac<0 \).
2. In the conic the coefficients of x², xy, y² are a, 2b, c, so the xy-coefficient is 2h with h = b, and the discriminant is \( h^2-ac=b^2-ac \).
3. Since \( b^2-ac<0 \), the conic is an ellipse.

Teacher's Note:
1. Compare the discriminant of the quadratic with h² − ac for the conic.
2. h² − ac < 0 means an ellipse.

Answer: (d) \( a\in(-1,5) \)

Explanation:
1. For an ellipse both denominators are positive, and the major axis is along the y-axis when the y² denominator is larger: \( f(a^2-5)>f(4a) \).
2. Since f is decreasing, this means \( a^2-5<4a \).
3. So \( a^2-4a-5<0 \), i.e. \( (a-5)(a+1)<0 \), giving \( a\in(-1,5) \).

Teacher's Note:
1. Major axis on the y-axis means the larger denominator is under y².
2. A decreasing function reverses inequalities.

Answer: (c) (0, 1)

Explanation:
1. A tangent to the parabola is \( y=mx+\frac am \). It touches the ellipse if \( \frac{a^2}{m^2}=a^2m^2+b^2 \), i.e. \( a^2m^4+b^2m^2-a^2=0 \).
2. Put \( u=m^2 \): \( g(u)=a^2u^2+b^2u-a^2 \) has \( g(0)=-a^2<0 \) and \( g(1)=b^2>0 \).
3. So the positive root u lies in (0, 1), hence \( m\in(0,1) \).

Teacher's Note:
1. Tangent y = mx + a/m must satisfy the ellipse tangency condition c² = a²m² + b².
2. Locate the positive root of the resulting quadratic in m².

Answer: (c) 1

Explanation:
1. A second-degree equation with no xy term is a rectangular hyperbola when the coefficients of \( x^2 \) and \( y^2 \) are equal in magnitude and opposite in sign, i.e. their sum is zero.
2. So \( (a-2)+a=0 \), giving a = 1.
3. Check: \( -x^2+y^2=4 \) is a rectangular hyperbola.

Teacher's Note:
1. Rectangular hyperbola: coefficient of x² + coefficient of y² = 0.
2. Check the result.

Answer: (c) \( \sqrt2 \)

Explanation:
1. The slopes of the asymptotes are \( -\frac34 \) and \( \frac43 \); their product is −1, so the asymptotes are perpendicular.
2. A hyperbola with perpendicular asymptotes is rectangular, and its eccentricity is \( \sqrt2 \).

Teacher's Note:
1. Check whether the asymptotes are perpendicular.
2. Eccentricity of a rectangular hyperbola.

Answer: (c) \( a^2 \)

Explanation:
1. The tangent at \( (a\sec\theta,a\tan\theta) \) is \( x\sec\theta-y\tan\theta=a \).
2. It meets y = x at \( P=(p,p) \) with \( p=\frac a{\sec\theta-\tan\theta} \) and y = −x at \( Q=(q,-q) \) with \( q=\frac a{\sec\theta+\tan\theta} \).
3. Area of triangle OPQ \( =\frac12|p(-q)-qp|=pq=\frac{a^2}{\sec^2\theta-\tan^2\theta}=a^2 \).

Teacher's Note:
1. Take the tangent in parametric form.
2. Find where it meets the two asymptotes.

Answer: (c) on the curve

Explanation:
1. Take the hyperbola \( xy=c^2 \) and the vertices \( \left(ct_i,\frac c{t_i}\right) \), i=1, 2, 3.
2. The altitude from the first vertex has slope \( t_2t_3 \), and the orthocentre works out to \( \left(-\frac c{t_1t_2t_3},\ -ct_1t_2t_3\right) \).
3. This point satisfies \( xy=c^2 \), so it lies on the curve.

Teacher's Note:
1. Use the parametric form (ct, c/t).
2. Substitute the orthocentre in xy = c².

Answer: (b) \( \frac\pi2 \)

Explanation:
1. The director circle of the hyperbola (locus of the points from which tangents are perpendicular) is \( x^2+y^2=a^2-b^2=25-16=9 \).
2. The given circle is exactly this director circle, so the tangents are perpendicular: angle \( \frac\pi2 \).

Teacher's Note:
1. Director circle of \( x^2/a^2-y^2/b^2=1 \) is x² + y² = a² − b².
2. Compare with the given circle.

Answer: (d) 8

Explanation:
1. Complete the squares: \( (x-2)^2-3(y+1)^2=11+4-3=12 \).
2. So \( \frac{(x-2)^2}{12}-\frac{(y+1)^2}{4}=1 \), with \( c^2=12+4=16 \), c = 4.
3. The distance between the foci is 2c = 8.

Teacher's Note:
1. Complete the squares in x and y.
2. Distance between foci = 2c, c² = a² + b².

Answer: (c) \( 2xy-4x+4y+1=0 \)

Explanation:
1. For a rectangular hyperbola \( e=\sqrt2 \). The focus-directrix property gives \( (x-1)^2+(y+1)^2=2\cdot\frac{(x-y+1)^2}2=(x-y+1)^2 \).
2. Expanding: \( x^2+y^2-2x+2y+2=x^2+y^2+1-2xy+2x-2y \).
3. Hence \( 2xy-4x+4y+1=0 \).

Teacher's Note:
1. Distance from the focus = e × distance from the directrix.
2. e = √2.

Answer: (a) 2

Explanation:
1. Take the point \( (h,h+c) \) on the line. Its chord of contact is \( \frac{hx}2-(h+c)y=1 \), i.e. \( h\left(\frac x2-y\right)=1+cy \).
2. For a fixed point to lie on this chord for every h: \( \frac x2-y=0 \) and \( 1+cy=0 \).
3. So \( y_1=-\frac1c \), \( x_1=2y_1 \), and \( \frac{x_1}{y_1}=2 \).

Teacher's Note:
1. Write the chord of contact T = 0 for (h, h + c).
2. Separate the terms containing h.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The ex-radii opposite B and C are \( r_2=\frac\Delta{s-b} \) and \( r_3=\frac\Delta{s-c} \), so \( \frac{r_2}{r_3}=\frac{s-c}{s-b}=k \) (constant).
2. This gives \( a+b-c=k(a+c-b) \), i.e. \( c-b=a\,\frac{1-k}{1+k} \), a constant since a = BC is fixed. So AB − AC is a constant smaller than BC.
3. The difference of the distances of A from the fixed points B and C is constant: A moves on a hyperbola with foci B, C. This is the defining property stated in R, so R explains A.

Teacher's Note:
1. Express r₂ and r₃ using Δ and the semi-perimeter.
2. A constant difference of distances from two fixed points defines a hyperbola.