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In solving this problem, I first solved the slope of the perpendicular bisector and I solved the midpoint of the two points. I don't know what to do next. I hope you can help me.

Lesley Murphy
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    One way to solve this problem is to assume the centre to be the point $(h,k)$ and then solve for $h$ and $k$ using the fact that both the given points lying on the circle are at a distance of given radius from the centre. You would get two equations. – Aman Kushwaha May 04 '22 at 11:32
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    For what it's worth, there are two distinct points $(a_1,b_1)$ and $(a_2,b_2)$, either of which may be the center of a satisfying circle. The easy way to see that is to notice that half the distance from $(1,-7)$ and $(5,9)$ is less than $\sqrt{85}.$ So, imagine the circular arc of radius $\sqrt{85}$ centered at $(1, -7)$ intersecting the circular arc of radius $\sqrt{85}$ centered at $(5,9)$ The two circular arcs will intersect at two distinct points, rather than only one point. – user2661923 May 04 '22 at 12:25
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    Does this answer your question? How to find center of a circle given 2 points on the circle and the radius? – mr_e_man May 04 '22 at 23:12

2 Answers2

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hint...let the circle centre be $(a,b)$

You require the line from the centre to the midpoint of the chord to be perpendicular to the chord, i.e $$\frac{b-1}{a-3}=-\frac14$$ and you need the distance from the centre to one of the points given to be equal to the radius, so $$(a-5)^2+(b-9)^2=85$$

Solve simultaneously.

David Quinn
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Let center to be $(a,b)$. $$(a-5)^2+(b-9)^2=85, (a-1)^2+(b+7)^2=85$$ $$a^2-10a+b^2-18b=-21,a^2-2a+b^2+14b=35$$ Subtracting gives $$8a+32b=56\Rightarrow a+4b=7\Rightarrow a=7-4b$$ $$35=a^2-2a+b^2+14b=49-56b+16b^2-14+8b+b^2+14b=35+17b^2-34b\Rightarrow b^2-2b=0$$ $$b=0\Rightarrow a=7-4b=7; b=2 \Rightarrow a=7-4b=-1$$ $$(x-7)^2+y^2=85 \mathrm{\ or\ } (x+1)^2+(y-2)^2=85$$

Ivan Kaznacheyeu
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