# Random point inside a circle

A point inside a circle of radius $$\sqrt{50}$$ lies $$2$$ units directly below a point on the circle, and $$6$$ units directly to the right of a point on the circle. What is the distance from the center of the circle to this point?

Can't find anything, this was rated as an easy question so i think there will be just one concept for this. I tried power of a point and and coordinate geometry but I cant answer it

## 1 Answer

Let the centre of the circle be at the origin, and let the point have coordinates $$(p,q)$$. Then $$p^2+(q+2)^2=50 \ \text{(eq. 1)}$$, and $$(p-6)^2+q^2=50 \ \text{(eq. 2)}$$, so $$p^2+(q+2)^2=(p-6)^2+q^2 \ \text{(eq. 3)}$$.

Expanding, we have:

$$p^2+q^2+4q+4=p^2-12p+36+q^2$$ $$4q+4=-12p+36$$ $$q+1=-3p+9$$ $$q=-3p+8 \ \text{(eq. 4)}$$

You can then substitute back into equation $$1$$ to find $$p$$. Then you can use equation $$2$$ and subtract $$-12p+36$$ from the LHS to find $$p^2+q^2$$, then $$\sqrt{p^2+q^2}$$.

• You might want to start off by saying, let the center of the circle be at the origin. – Gerry Myerson Nov 16 '18 at 11:51
• @GerryMyerson Fixed. – Toby Mak Nov 16 '18 at 11:52