find the equation of the circle which passes through the centre of the circle x^2+y^2-4x-8y-41=0 and is concentric with the circle x^2+y^2-2y+1=0
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I solved the question,but there is something weird since the radius of second given circle is zero.
But still I solved it
Hope it helps...
:)
But still I solved it
Hope it helps...
:)
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Ridhikamini:
thank you..but the answer isn't right. I think there maybe some mistake in the problem
ya cuz radius cant be zero
thats what i mentioned
the answer is right
thank you
ok np
so answer is correct or not?
ans is correct
ok
Answered by
6
- Given : x²+y²-4x-8y-41=0
- X coordinate of center =(-Coefficient of x/2).
- Y coordinate of center =(-Coefficient of y/2).
- So the center of the circle is (2,4).
⇒x²+y²-2y+1=0
⇒(x)²+(y-1)²=0 (This is a point circle with center (0,1)
- Let the equation of the circle be (x)²+(y-1)²=r²
- Point (0,1) should satisfy this equation.
⇒ (2)²+(4-1)²=r²
⇒4+9=r²
⇒r²=13
∴The equation of the required circle is (x)²+(y-1)²=13
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