Question

The least and the greatest distance of the point $(10,7)$ from the circle $x^2+y^2+4x-2y-20=0$ are

• A. $10,5$
• B. $15,20$
• C. $12,16$
• D. $5,15$

Answer 1

The correct option is D

$5,15$

Solve the given expression to find the distance

Given expression, $x^2+y^2+4x-2y-20=0$

adding $(4+1)$ on both sides and applying the formula ${(a-b)}^2=a^2-2ab+b^2$ we get,

$\Rightarrow {(x-2)}^2+{(y-1)}^2=20+4+1$

$\Rightarrow (x-2)²+(y-1)²=25$

$\Rightarrow (x-2)²+(y-1)²=5²$

So, the circle's centre is at point $C$ $(2,1)$ and radius $r=5$,

Now, the distance between point $C$ $(2,1)$ and $P$ $(10,7)$ is

$$\begin{gathered} =\sqrt{{(10-2)}^2+{(7-1)}^2} \\ =\sqrt{8^2+6^2} \\ =\sqrt{64+36} \\ =\sqrt{100} \\ =10 \end{gathered}$$

So, the radius is less than the distance between the points and $P(10,7)$ lies outside the circle.

Now, the least distance will be $10-5=5$ units and, the greatest distance will be $10+5=15$ units.

Hence, the correct answer is option (D).