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If a circle C, whose radius is 3, touches externally the circle,
$x^{2}+y^{2}+2x-47y-4=0$ at the point (2, 2), then the length of the intercept cut by this
circle C, on the x-axis is equal to :

Option 1)
$2\sqrt{5}$

Option 2)
$3\sqrt2$

Option 3)
$\sqrt5$

Option 4)
$2\sqrt3$

Answers (1)

best_answer

As we have learned

Selection formula -

$x= \frac{mx_{2}+nx_{1}}{m+n}$

$y= \frac{my_{2}+ny_{1}}{m+n}$

- wherein

If P(x,y) divides the line joining A(x₁,y₁) and B(x₂,y₂) in ration $m:n$

General form of a circle -

$x^{2}+y^{2}+2gx+2fy+c= 0$

- wherein

centre = $\left ( -g,-f \right )$

radius = $\sqrt{g^{2}+f^{2}-c}$

$S= x^{2}+y^{2}+2gx+2fy+c=0$

equation of

$C_{1} is (x-5)^{2}+(y-2)^{2}=3^{2}$

$x^{2}+y^{2}-10x-4y+20 =0$

X- intercept = $2\sqrt (g^{2}-c)= 2\sqrt(25-20)$

$= 2\sqrt5$

Option 1)

$2\sqrt{5}$

This is correct

Option 2)

$3\sqrt2$

This is incorrect

Option 3)

$\sqrt5$

This is incorrect

Option 4)

$2\sqrt3$

This is incorrect

Posted by

Himanshu

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