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(C) Let the center of the circle be $(h, k)$. Since the center lies on the line $x + y = 4$,we have $h + k = 4$,so $k = 4 - h$.Since the circle passes

Vedclass · Accepted Answer

$x^2 + y^2 - 4x - 4y = 0$

Vedclass · Answer

(C) Let the center of the circle be $(h, k)$. Since the center lies on the line $x + y = 4$,we have $h + k = 4$,so $k = 4 - h$.Since the circle passes through the origin $(0, 0)$,its equation is $x^2 + y^2 - 2hx - 2ky = 0$.Substituting $k = 4 - h$,the equation becomes $x^2 + y^2 - 2hx - 2(4 - h)y = 0$.Two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ intersect orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = -h, f_1 = -(4 - h), c_1 = 0$ and $g_2 = -2, f_2 = 1, c_2 = 4$.Applying the condition: $2(-h)(-2) + 2(-(4 - h))(1) = 0 + 4$.$4h - 8 + 2h = 4$.$6h = 12$,which gives $h = 2$.Then $k = 4 - 2 = 2$.The equation of the circle is $x^2 + y^2 - 2(2)x - 2(2)y = 0$,which simplifies to $x^2 + y^2 - 4x - 4y = 0$.

The equation of the circle passing through the origin,having its center on the line $x + y = 4$,and intersecting the circle $x^2 + y^2 - 4x + 2y + 4 = 0$ orthogonally is:

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