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Algebra> a circle passes through the point(0,0)And...

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a circle passes through the point(0,0)And(1,0) and touch the circle internally x²+y²=9. find the equation of the circle.

ashish , 7 Years ago

Grade 11

1 Answers

Ritesh Khatri

Last Activity: 7 Years ago

Let the equation be $x^{2}+y^{2}+2gx+2fy+c=0$

Since it pass through (0,0) c=0

there fore equation becomes $x^{2}+y^{2}+2gx+2fy=0$

since this pass through (1,0)

$1^{2}+0^{2}+2g1+2f0=0$

$\Rightarrow 1+2g=0$

$\Rightarrow g=\frac{-1}{2}$

equation becomes $x^{2}+y^{2}-x+2fy=0$

centre = (1/2 , -f)

radius = [(-1/2)^2 + (-f)^2]^(1/2)

= [(1/4)+f^2]^(1/2)

radius of given circle is 3

since circle touch internally |r₁ – r₂| = C₁C₂

which gives f = 2^(1/2)

hence equation of circle is $x^{2}+y^{2}-x-2\sqrt{2} y=0$

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