10-1.Circle and System of Circles
hard

જો વર્તુળ $C$ જેની ત્રિજ્યા $3$ હોય તે વર્તુળ $x^2 + y^2 + 2x - 4y - 4 = 0$ ને બહારથી બિંદુ $(2, 2)$ આગળ સ્પર્શે તો વર્તુળ $C$ એ $x-$ અક્ષ સાથે બનાવેલ અંત:ખંડની લંબાઈ  મેળવો. 

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

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