Question

Find the locus of the mid point of the chord of a circle $x^2 + y^2 = 4$ such that the segment intercepted by the chord on the curve $x^2 - 2x - 2y = 0$ subtends a right angle at the origin.

• A. (x-1)^2 + (y-1)^2 = 2
• B. x^2 + y^2 = 2
• C. (x-1)^2 + (y-1)^2 = 4
• D. x^2 + y^2 = 4

Answer 1

Answer: (A)

(x-1)^2 + (y-1)^2 = 2

Answer 2

Let $M(h,k)$ be the midpoint of a chord of the circle $C_1: x^2 + y^2 = 4$. The equation of the chord is $hx + ky = h^2 + k^2$. Let this line be $L$.

The line $L$ intersects the parabola $C_2: x^2 - 2x - 2y = 0$ at two points, $P(x_1, y_1)$ and $Q(x_2, y_2)$. The condition is that $\angle POQ = 90^\circ$, which means $x_1x_2 + y_1y_2 = 0$.

From $L$, $y = -\frac{h}{k}x + \frac{h^2+k^2}{k}$ (assuming $k \neq 0$). Substituting this into the parabola equation: $x^2 - 2x - 2\left(-\frac{h}{k}x + \frac{h^2+k^2}{k}\right) = 0$ $kx^2 + (2h - 2k)x - 2(h^2+k^2) = 0$ By Vieta's formulas, $x_1x_2 = \frac{-2(h^2+k^2)}{k}$.

Since $P$ and $Q$ lie on the parabola, $y = \frac{x^2-2x}{2}$. So, $y_1y_2 = \frac{1}{4}(x_1^2-2x_1)(x_2^2-2x_2) = \frac{1}{4}((x_1x_2)^2 - 2x_1x_2(x_1+x_2) + 4x_1x_2)$. Also, $x_1+x_2 = \frac{2k-2h}{k}$. Substituting these into the expression for $y_1y_2$: $y_1y_2 = \frac{1}{4}\left(\left(\frac{-2(h^2+k^2)}{k}\right)^2 - 2\left(\frac{-2(h^2+k^2)}{k}\right)\left(\frac{2k-2h}{k}\right) + 4\left(\frac{-2(h^2+k^2)}{k}\right)\right)$ $y_1y_2 = \frac{h^2+k^2}{k^2}(h^2+k^2 - 2h)$

Now, applying $x_1x_2 + y_1y_2 = 0$: $\frac{-2(h^2+k^2)}{k} + \frac{h^2+k^2}{k^2}(h^2+k^2 - 2h) = 0$ $\frac{h^2+k^2}{k}\left(-2 + \frac{h^2+k^2-2h}{k}\right) = 0$ This yields $-2k + h^2+k^2-2h = 0$, or $h^2 - 2h + k^2 - 2k = 0$.

Replacing $h$ with $x$ and $k$ with $y$, we get $x^2 - 2x + y^2 - 2y = 0$. Completing the square: $(x-1)^2 + (y-1)^2 = 2$. This is the equation of a circle with center $(1,1)$ and radius $\sqrt{2}$. The case $k=0$ needs to be checked. If $k=0$, the midpoint is $(h,0)$ and the chord is $x=h$. This line intersects the parabola at a single point, so no segment is intercepted. Thus $k \neq 0$ is required. The origin $(0,0)$ is not a solution because the angle $\angle POQ$ is undefined if $P=O$. The derived locus $(x-1)^2 + (y-1)^2 = 2$ does not contain the origin.