Given: Radius of circle is 3
Radius of other circle = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{(-1)^2+2^2-(-4)}=3\)
Now, circle C is touching externally other circle at point (2,2)
Therefore, (2,2) is midpoint of two centres of the circles as radius are equal.
Let the coordinate of centre of circle C be (p,q).
Centre of other circle is (−1,2)
⇒ \(\frac{p-1}{2}\) = and \(\frac{q+2}{2}\) = 2
⇒ p = 5 and q = 2
Equation of circle C is
(x − 5)2 + (y − 2)2 =32
⇒ x2 + y2 −10x − 4y + 20 = 0
x-intercept = \(2\sqrt{g^2-c}\)
Here, g = (−5) and c = 20
x-intercept = \(2\sqrt{(-5)^2-20}\)
= 2√5