If P(6,1) is the orthocentre of the triangle | vedclass

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(A) Let $P(6,1)$ be the orthocentre of $	riangle ABC$ with $A(5,-2)$,$B(8,3)$,and $C(h, k)$.Since $AP \perp BC$,the slope of $AP = \frac{1 - (-2)}{6

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$x^2+y^2-65=0$

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(A) Let $P(6,1)$ be the orthocentre of $	riangle ABC$ with $A(5,-2)$,$B(8,3)$,and $C(h, k)$.Since $AP \perp BC$,the slope of $AP = \frac{1 - (-2)}{6 - 5} = \frac{3}{1} = 3$.Therefore,the slope of $BC = -\frac{1}{3}$.The equation of line $BC$ passing through $B(8,3)$ is $y - 3 = -\frac{1}{3}(x - 8)$,which simplifies to $3y - 9 = -x + 8$,or $x + 3y - 17 = 0$.Since $BP \perp AC$,the slope of $BP = \frac{1 - 3}{6 - 8} = \frac{-2}{-2} = 1$.Therefore,the slope of $AC = -1$.The equation of line $AC$ passing through $A(5,-2)$ is $y - (-2) = -1(x - 5)$,which simplifies to $y + 2 = -x + 5$,or $x + y - 3 = 0$.To find $C(h, k)$,we solve the system:$1) x + 3y = 17$$2) x + y = 3$Subtracting $(2)$ from $(1)$ gives $2y = 14$,so $y = 7$.Substituting $y = 7$ into $(2)$ gives $x + 7 = 3$,so $x = -4$.Thus,$C = (-4, 7)$.Now,check which circle equation is satisfied by $C(-4, 7)$:$(-4)^2 + (7)^2 = 16 + 49 = 65$.So,$x^2 + y^2 - 65 = 0$ is the correct equation.

If $P(6,1)$ is the orthocentre of the triangle whose vertices are $A(5,-2)$,$B(8,3)$,and $C(h, k)$,then the point $C$ lies on the circle:

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