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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
anser 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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