How to prove that the circle $x^2+y^2−6x−4y+9=0$ bisects the circumference of the circle $x^2+y^2−8x−6y+23=0$ ?
How to prove a circle bisects circumference of another circle
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analytic-geometry
2 Answers
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Equation of common chord: $$-6x-4y+9=-8x-6y+23$$
$$x+y=7$$
Centre of the second circle:
$$(4,3)$$
which lies on the common chord.
Hence the common chord is the diameter of the second circle.
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Solving simultaneously, the intersection points lie on the line $x+y=7$. The centre of the second circle is $(4,3)$ which lies on this line.