Consider a circle (x- )^2+(y- )^2=50, where , 0.

Question

Consider a circle $(x-\alpha {)}^{2}+(y-\beta {)}^{2}=50$, where $\alpha ,\beta >0$. If the circle touches the line $y+x=0$ at the point $P$, whose distance from the origin is $4\sqrt{2}$, then $(\alpha +\beta {)}^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 100

Accepted Answer

Radius $=\sqrt{50}=5\sqrt2$. In the right triangle formed by the centre $C=(\alpha,\beta)$, the origin $O$, and the tangent point $P$ (with $CP\perp OP$ since $CP$ is a radius to the tangent line):

$$OC^2=OP^2+CP^2=(4\sqrt2)^2+(5\sqrt2)^2=32+50=82.$$

Alternatively, since the line touches the circle, the distance from $C$ to $x+y=0$ equals the radius:

$$\frac{|\alpha+\beta|}{\sqrt2}=5\sqrt2 \;\Rightarrow\; \alpha+\beta=10 \;\Rightarrow\; (\alpha+\beta)^2=100.$$

Solutions · 1