Let a circle C:(x-h)^2+(y-k)^2=r^2, k0, touch the x-axis at (1,0).

Question

Let a circle $C:(x-h)^2+(y-k)^2=r^2$, $k>0$, touch the $x$-axis at $(1,0)$. If the line $x+y=0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $PQ$ is $2$, then the value of $h+k+r$ is equal to:

Answer: 7

Accepted Answer

Tangency to the $x$-axis at $(1,0)$ gives $h=1,\ r=k$.

Distance from centre to $x+y=0$: $d=\dfrac{|1+k|}{\sqrt2}$. Half-chord $=1$:
$$k^2-\frac{(1+k)^2}2=1\ \Longrightarrow\ k^2-2k-3=0\ \Longrightarrow\ k=3\ (k>0).$$

$h+k+r=1+3+3=7.$

Solutions · 1