Jamia Millia Islamia MCA Circle Previous Year Questions (PYQs)

Median of equilateral triangle = $\dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow 3a = \dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow \text{side} = 2\sqrt{3}a$ Radius (distance from centre to vertex) = $\dfrac{\text{side}}{\sqrt{3}} = 2a$. Equation: $x^2 + y^2 = (2a)^2 = 4a^2$. $\boxed{\text{Answer: (C) }x^2 + y^2 = 4a^2}$

**Solution:** Let $P(x_1, y_1)$ be a point on $x^2 + y^2 = 1$. Midpoint $M$ of $AP$ is \[ \left(\dfrac{x_1 + 3}{2}, \dfrac{y_1 + 2}{2}\right) \Rightarrow x_1 = 2x - 3,\ y_1 = 2y - 2. \] Since $P$ lies on the circle: \[ (2x - 3)^2 + (2y - 2)^2 = 1 \] \[ \Rightarrow 4x^2 + 4y^2 - 12x - 8y + 13 = 0 \] Divide by 4: \[ x^2 + y^2 - 3x - 2y + \dfrac{13}{4} = 0 \] \[ \Rightarrow (x - \tfrac{3}{2})^2 + (y - 1)^2 = \dfrac{11}{4} \] Hence radius $= \dfrac{\sqrt{11}}{2}$.

For ellipse $4x^2 + 9y^2 = 36 \Rightarrow \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ and for circle $4x^2 + 4y^2 = 31 \Rightarrow x^2 + y^2 = \dfrac{31}{4}$ Let tangent be $y = mx + c$. For ellipse, condition is $c^2 = a^2m^2 + b^2 = 9m^2 + 4$. For circle, condition is $c^2 = r^2(1 + m^2) = \dfrac{31}{4}(1 + m^2)$. Equating both: $9m^2 + 4 = \dfrac{31}{4}(1 + m^2)$ $\Rightarrow 36m^2 + 16 = 31 + 31m^2$ $\Rightarrow 5m^2 = 15 \Rightarrow m^2 = 3.$

Equation of circle: $(x - h)^2 + (y - k)^2 = r^2$ Substitute $(h, k) = (0, -2)$ and $r = 2$: $x^2 + (y + 2)^2 = 4 \Rightarrow x^2 + y^2 + 4y = 0.$

Equation of chord of contact from $(a\cos\theta, a\sin\theta)$ to circle $x^2 + y^2 = b^2$ is $a(\cos\theta \,x + \sin\theta \,y) = b^2.$ For this line to be tangent to $x^2 + y^2 = c^2$, the perpendicular distance from origin = radius of that circle. $\Rightarrow \dfrac{|b^2|}{\sqrt{a^2}} = c$ $\Rightarrow b^2 = a c$ Hence, $a, b, c$ are in G.P.