The straight line x+y-1=0 meets the circle x^ | vedclass

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(A) The equation of any circle passing through the intersection of the circle $S: x^2+y^2-6x-8y=0$ and the line $L: x+y-1=0$ is given by $S + \lambda

Vedclass · Accepted Answer

$x^2+y^2-2y-6=0$

Vedclass · Answer

(A) The equation of any circle passing through the intersection of the circle $S: x^2+y^2-6x-8y=0$ and the line $L: x+y-1=0$ is given by $S + \lambda L = 0$. $x^2+y^2-6x-8y + \lambda(x+y-1) = 0$ $x^2+y^2 + (\lambda-6)x + (\lambda-8)y - \lambda = 0$. Since $AB$ is a diameter, the center of this circle must lie on the line $x+y-1=0$. The center of the circle is $(-\frac{\lambda-6}{2}, -\frac{\lambda-8}{2})$. Substituting the center into the line equation: $-\frac{\lambda-6}{2} - \frac{\lambda-8}{2} - 1 = 0$ $-(\lambda-6) - (\lambda-8) - 2 = 0$ $-\lambda + 6 - \lambda + 8 - 2 = 0$ $-2\lambda + 12 = 0 \implies \lambda = 6$. Substituting $\lambda = 6$ into the circle equation: $x^2+y^2 + (6-6)x + (6-8)y - 6 = 0$ $x^2+y^2-2y-6=0$.

The straight line $x+y-1=0$ meets the circle $x^2+y^2-6x-8y=0$ at $A$ and $B$. Then the equation of the circle of which $AB$ is a diameter is

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