Question

What is the maximum and minimum distance of the point (9, 12) from the circle $x^2+y^2-6x-8y-24=0$

• A. 15 and 5
• B. 14 and 6
• C. 12 and 8
• D. 17 and 3

Answer 1

The correct option is D

17 and 3

Before we draw the diagram of circle and the point, we have to check it it lies outside of inside the

circle.

For that we will substitute the point in the equation of the circle and find $S_1$

$S_1>0\Rightarrow$ it lies outside,

$S_1=0\Rightarrow$ lies on the circle

$S_1<0\Rightarrow$ lies inside the circle

$S_1=9^2+{12}^2-6\times 9-8\times 12-24=51>0$

$\Rightarrow$ The point lies outside the circle

Let the farthest point be A and nearest point be B. instead of calculating PA and PB directly, we will find

PC first and then use the relation.

$PA=PC+AC=PC+r$

And

$PB=PC-CB=PC-r$

For that we will find the radius and centre of $x^2+y^2-6x-8y-24=0$

Centre $C\equiv (3,4)$ and radius $r=\sqrt{3^2+4^2-(-24)}$

$=7$

$\Rightarrow PC=\sqrt{(9-3{)}^2+(12-4{)}^2}$

$=10$

$\Rightarrow PA=10+7=17$ and

$PB=10-7=3$