Find the coordinates of the centre and the radius of the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-8=0$. I

Question

Find the coordinates of the centre and the radius of the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-8=0$. If the circle cuts the $x$-axis at $P$ and $Q$, find the equation of the perpendicular bisector of the line $PQ$. Given that the line $2y-3x+7=0$ intersects the circle at $R$ and $S$, find the coordinates of $R$ and $S$. Hence, show that the line segment $RS$ is a diameter of the circle.

Accepted Answer

(a) Centre $(1,-2)$, radius $\sqrt{13}$ (b) $x=1$ (c) $R=(3,1)$, $S=(-1,-5)$, and their midpoint is $(1,-2)$, so $RS$ is a diameter

Coordinate Geometry - Line–Circle Intersections

Coordinate Geometry - Line–Circle Intersections

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