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(A) The given circle is $S: x^2 + y^2 - 16 = 0$. The equation of the chord is $L: 3x + y + 5 = 0$. The equation of the family of circles passing throu

Vedclass · Accepted Answer

$x^2 + y^2 + 3x + y - 11 = 0$

Vedclass · Answer

(A) The given circle is $S: x^2 + y^2 - 16 = 0$. The equation of the chord is $L: 3x + y + 5 = 0$. The equation of the family of circles passing through the intersection of $S$ and $L$ is $S + \lambda L = 0$. Thus,$x^2 + y^2 - 16 + \lambda(3x + y + 5) = 0$. $x^2 + y^2 + 3\lambda x + \lambda y + (5\lambda - 16) = 0$. The center of this circle is $C = (-\frac{3\lambda}{2}, -\frac{\lambda}{2})$. Since the chord $3x + y + 5 = 0$ is the diameter of this circle,the center $C$ must lie on the line $3x + y + 5 = 0$. Substituting the center into the line equation: $3(-\frac{3\lambda}{2}) + (-\frac{\lambda}{2}) + 5 = 0$. $-\frac{9\lambda}{2} - \frac{\lambda}{2} + 5 = 0$. $-5\lambda + 5 = 0 \implies \lambda = 1$. Substituting $\lambda = 1$ into the circle equation: $x^2 + y^2 + 3(1)x + (1)y + 5(1) - 16 = 0$. $x^2 + y^2 + 3x + y - 11 = 0$.

The equation of the circle described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as diameter is:

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