A circle x^2 + y^2 + 2gx + 2fy + c = 0 passin | vedclass

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(A) Two circles are concentric if they have the same center. The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The reference circle is $x^2 + y^2 -

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(A) Two circles are concentric if they have the same center. The given circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The reference circle is $x^2 + y^2 - 2x + 4y + 20 = 0$. Comparing the reference circle with $x^2 + y^2 + 2gx + 2fy + c = 0$,the center is $(-g, -f)$. For the reference circle,the center is $(-(-1), -(2)) = (1, -2)$. Thus,for the required circle,$g = -1$ and $f = 2$. The equation becomes $x^2 + y^2 - 2x + 4y + c = 0$. Since the circle passes through $(4, -2)$,we substitute these coordinates into the equation: $(4)^2 + (-2)^2 - 2(4) + 4(-2) + c = 0$ $16 + 4 - 8 - 8 + c = 0$ $4 + c = 0$ $c = -4$.

$A$ circle $x^2 + y^2 + 2gx + 2fy + c = 0$ passing through $(4, -2)$ is concentric to the circle $x^2 + y^2 - 2x + 4y + 20 = 0$. Then the value of $c$ is:

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