Let a circle C have its centre in the first quadrant, pass through the origin (so it meets…

Question

Let a circle $C$ have its centre in the first quadrant, pass through the origin (so it meets the coordinate axes at exactly three points), and cut off equal intercepts from the coordinate axes. If the length of the chord of $C$ on the line $x+y=1$ is $\sqrt{14}$, then the square of the radius of $C$ is \_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

Equal, equal-sign intercepts and centre in the first quadrant force the centre onto $y=x$: centre $(a,a)$, radius $r=a\sqrt2$ (since the circle $x^2+y^2-2ax-2ay=0$ has $x$-intercepts $0,2a$ and $y$-intercepts $0,2a$).

Distance from $(a,a)$ to $x+y-1=0$: $d=\dfrac{|2a-1|}{\sqrt2}$. Using $\left(\dfrac{\sqrt{14}}{2}\right)^2+d^2=r^2$:

$$\frac{14}{4}+\frac{(2a-1)^2}{2}=2a^2 \;\Rightarrow\; 14+2(2a-1)^2=8a^2 \;\Rightarrow\; 16-8a=0 \;\Rightarrow\; a=2.$$

$r^2=2a^2=8.$

Solutions · 1