A circle S given by x^2+y^2-14x+6y+33=0 cuts | vedclass

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(A) The circle $S$ is $x^2+y^2-14x+6y+33=0$. To find the intersection with the $X$-axis,set $y=0$: $x^2-14x+33=0 \implies (x-3)(x-11)=0$. Thus,the poi

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$x^2+y^2-17x+3y+54=0$

Vedclass · Answer

(A) The circle $S$ is $x^2+y^2-14x+6y+33=0$. To find the intersection with the $X$-axis,set $y=0$: $x^2-14x+33=0 \implies (x-3)(x-11)=0$. Thus,the points are $A(3, 0)$ and $B(11, 0)$ since $OB > OA$. The midpoint $C$ of $AB$ is $(\frac{3+11}{2}, 0) = (7, 0)$. The line $L$ passes through $(7, 0)$ with slope $-1$: $y-0 = -1(x-7) \implies x+y-7=0$. Since $L$ is the radical axis of $S$ and $S^{\prime}$,the equation of $S^{\prime}$ is $S + kL = 0$: $x^2+y^2-14x+6y+33 + k(x+y-7) = 0$. $x^2+y^2+(k-14)x+(k+6)y+(33-7k) = 0$. The center of $S^{\prime}$ is $(-\frac{k-14}{2}, -\frac{k+6}{2})$. Since $L$ is the diameter of $S^{\prime}$,the center must lie on $L$: $-\frac{k-14}{2} - \frac{k+6}{2} - 7 = 0 \implies -k+14-k-6-14 = 0 \implies -2k-6=0 \implies k=-3$. Substituting $k=-3$ into the equation: $x^2+y^2+(-3-14)x+(-3+6)y+(33-7(-3)) = 0$. $x^2+y^2-17x+3y+54=0$.

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