Question

A circle meets coordinate axes at 3 points and cuts equal intercepts. If it cuts a chord of length $\sqrt{14}$ unit on $x + y = 1$, then square of its radius is (centre lies in first quadrant):

• A. 2
• B. 4
• C. 8
• D. 16

Answer 1

The Correct Option is C

Step 1: Determine the circle's properties from intercepts.
The circle meets axes at 3 points and cuts equal intercepts. This implies the circle passes through the origin $(0,0)$ and two other points $(a, 0)$ and $(0, a)$.
The equation of such a circle is $x^2 + y^2 - ax - ay = 0$.
The centre is $(h, k) = (a/2, a/2)$ and the radius squared is $r^2 = (a/2)^2 + (a/2)^2 = a^2/2$.


Step 2: Use the chord length condition.
The circle cuts a chord of length $L = \sqrt{14}$ on the line $x + y - 1 = 0$.
The formula for chord length is $L = 2\sqrt{r^2 - d^2}$, where $d$ is the perpendicular distance from the centre to the line.


Step 3: Calculate $d$.
Distance from $(\frac{a}{2}, \frac{a}{2})$ to $x + y - 1 = 0$:
$d = \frac{|\frac{a}{2} + \frac{a}{2} - 1|}{\sqrt{1^2 + 1^2}} = \frac{|a - 1|}{\sqrt{2}}$
So, $d^2 = \frac{(a - 1)^2}{2}$.


Step 4: Solve for $a$.
$\sqrt{14} = 2\sqrt{\frac{a^2}{2} - \frac{(a-1)^2}{2}}$
Square both sides:
$14 = 4 \left( \frac{a^2 - (a^2 - 2a + 1)}{2} \right)$
$14 = 2(2a - 1)$
$7 = 2a - 1 \implies 2a = 8 \implies a = 4$


Step 5: Find the square of the radius.
$r^2 = a^2 / 2 = 4^2 / 2 = 16 / 2 = 8$.
The answer is 8 (Option 3).