The circle $x^2 + y^2 - 2x + 7y - 8 = 0$ cuts the $x$-axis at $A$ and $B$ and the $y$-axis

Question

The circle $x^2 + y^2 - 2x + 7y - 8 = 0$ cuts the $x$-axis at $A$ and $B$ and the $y$-axis at $C$ and $D$. Show that $OA \times OB = OC \times OD$, where $O$ is the origin. Show that $OA^2 + OB^2 + OC^2 + OD^2 = 4r^2$, where $r$ is the radius of the circle. Show that $AD \times BC + AC \times BD = AB \times CD$.

Accepted Answer

(i) $OA\times OB=OC\times OD=8$ (shown). (ii) $OA^2+OB^2+OC^2+OD^2=85=4r^2$ (shown). (iii) $AD\times BC+AC\times BD=54=AB\times CD$ (shown).

Coordinate Geometry - Line–Circle Intersections

Coordinate Geometry - Line–Circle Intersections

Solution:

Solution locked