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10-1.Circle and System of Circles
hard
यदि एक वृत्त $C$, जिसकी त्रिज्या 3 है, एक अन्य वृत्त $x^{2}+y^{2}+2 x-4 y-4=0$ को बाह्य रूप से बिंदु $(2,2)$ पर स्पर्श करता है, तो वृत्त $C$ द्वारा $x$-अक्ष पर काटे गए अंतःखंड की लंबाई है
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
