Question

The point P(9, 12) lies inside the circle $x^2+y^2-6x-8y-24=0$

• A. True
• B. False

Answer 1

The correct option is B

False

To find if a point lies outside the circle, we just have to check if its distance from center is more than

radius.

Centre $\equiv$ (3,4)

Radius = $\sqrt{3^2+4^2-(-24)}$

=7

Distance of P(9,12) from (3,4)

$=\sqrt{(9-3{)}^2+(12-4{)}^2}=10$, which is greater than the radius.

$\Rightarrow$ P lies outside the circle.

We can also find $S_1$ and check for its sign.

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

$S_1>0\Rightarrow P$ lies outside the circle