Question

If two circles ${(x-1)}^2+{(y-3)}^2=r^2$ and $x^2+y^2-8x+2y+8=0$ intersect in two distinct points, then

• A. $r<2$
• B. $8<r<10$
• C. $r=2$
• D. $2<r<8$

Answer 1

Answer: (D)

$2<r<8$

Let $x$ be the radius of the circle, so $x=r$,

The center of one circle is $(4,-1)$ with radius $3$, and of other circle is $(1,3)$ with radius $x$.

Distance between the centres $=\sqrt{(4-1{)}^2+(-1-3{)}^2}=\sqrt{9+16}=5$.

There are two extreme cases in the diagram, we have to find $x$ for intersection btw them.

from first figure, $x+3>5\to x>2$

And from second figure, $x<5+3\to x<8$

We have, $2<x<8$ i.e. $2<r<8$

Hence, D is correct.