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10-1.Circle and System of Circles
hard
If a circle $C,$ whose radius is $3,$ touches externally the circle, $x^2 + y^2 + 2x - 4y - 4 = 0$ at the point $(2, 2),$ then the length of the intercept cut by circle $c,$ on the $x-$ axis is equal to
A
$\sqrt 5$
B
$2\sqrt 3$
C
$3\sqrt 2$
D
$2\sqrt 5$
(JEE MAIN-2018)
Solution
Given circle is:
${x^2} + {y^2} + 2x – 4y – 4 = 0$
$\therefore $ it center is $\left( { – 1,2} \right)$ and radius is $3$ units.
Let $A=(x,y)$ be the center of the circle $C$
$\therefore \frac{{x – 1}}{2} = 2 \Rightarrow x = 5$
$ \Rightarrow y = 2$
So the center of $C$ is $(5,2)$ and its radius is $3$
$\therefore $ equation of center $C$ is:
${x^2} + {y^2} – 10x – 4y + 20 = 0$
$\therefore $ The length of the intercept it cuts on the $X$-axis
$ = 2\sqrt {{g^2} – c} = 2\sqrt {25 – 20} = 2\sqrt 5 $
Standard 11
Mathematics
