Question

The point on the circle $x^2+y^2-6x+4y-12=0$ which is at maximum distance from the point $(-9,7)$ is

• A. $(-1,1)$
• B. $(7,-5)$
• C. $(0,-6)$
• D. $(0,2)$

Answer 1

The correct option is B $(7,-5)$

The points on maximum and minimum distance from (-9,7) must lie on normal from it to the circle.

Circle is $(x-3{)}^2+(y+2{)}^2=25$

Centre of circle is $(3,-2)$

Equation of normal passing through $(-9,7)$ and $(3,-2)$ is $3x+4y-1=0$

Intersection of normal and circle gives two points which are $(-1,1)$ and $(7,-5)$.

Clearly, $(7,-5)$ is at maximum distance from $(-9,7)$.

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