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If $S=x^2+y^2+2 x+17 y+4=0$
$S^{\prime}=x^2+y^2+7 x+6 y+11=0$ and
$S^{\prime \prime}=x^2+y^2-x+22 y+3=0$ are three
circles, then the length of tangent from their radical center to $S=0$ is ......... units.
Options:
$S^{\prime}=x^2+y^2+7 x+6 y+11=0$ and
$S^{\prime \prime}=x^2+y^2-x+22 y+3=0$ are three
circles, then the length of tangent from their radical center to $S=0$ is ......... units.
Solution:
2936 Upvotes
Verified Answer
The correct answer is:
$\sqrt{57}$
Given, $S=x^2+y^2+2 x+17 y+4=0$
$S^{\prime \prime}=x^2+y^2+7 x+6 y+11=0$
$S=x^2+y^2-x+22 y+3=0$
$S-S^{\prime}=5 x-11 y+7=0$ ...(i)
$S^{\prime}-S^{\prime \prime}=8 x-16 y+8=0$
$x-2 y+1=0$ ...(ii)
Solving Eqs. (i) and (ii), we get $(3,2)$
$\therefore$ Radical centre of circle is $(3,2)$.
Length of tanget from $(3,2)$ to $S$ is $\sqrt{S_1}$
$=\sqrt{(3)^2+(2)^2+2(3)+17(2)+4}=\sqrt{57}$
$S^{\prime \prime}=x^2+y^2+7 x+6 y+11=0$
$S=x^2+y^2-x+22 y+3=0$
$S-S^{\prime}=5 x-11 y+7=0$ ...(i)
$S^{\prime}-S^{\prime \prime}=8 x-16 y+8=0$
$x-2 y+1=0$ ...(ii)
Solving Eqs. (i) and (ii), we get $(3,2)$
$\therefore$ Radical centre of circle is $(3,2)$.
Length of tanget from $(3,2)$ to $S$ is $\sqrt{S_1}$
$=\sqrt{(3)^2+(2)^2+2(3)+17(2)+4}=\sqrt{57}$
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