Let P be a moving point on the circle x^2+y^2-6 x-8 y+21=0.

Question

Let P be a moving point on the circle $x^2+y^2-6 x-8 y+21=0$. Then, the maximum distance of P from the vertex of the parabola $x^2+6 x+y+13=0$ is equal to:

Accepted Answer

First, find the vertex of the parabola.

The parabola is

${x}^{2}+6x+y+13=0$

Rewrite it by completing the square:

${x}^{2}+6x+y+13=0$

${x}^{2}+6x+9+y+13-9=0$

$(x+3{)}^{2}+y+4=0$

$y=-(x+3{)}^{2}-4$

So, the vertex of the parabola is

$(-3,-4)$

Now, find the centre and radius of the circle.

Given circle:

${x}^{2}+{y}^{2}-6x-8y+21=0$

Complete the squares:

${x}^{2}-6x+{y}^{2}-8y+21=0$

$({x}^{2}-6x+9)+({y}^{2}-8y+16)+21-9-16=0$

$(x-3{)}^{2}+(y-4{)}^{2}-4=0$

$(x-3{)}^{2}+(y-4{)}^{2}=4$

Hence, centre of the circle is

$(3,4)$

and radius is

$r=2$

Now we need the maximum distance of a point $P$ on this circle from the vertex $(-3,-4)$.

Let $V=(-3,-4)$ and centre of circle be $C=(3,4)$.

First find $VC$:

$VC=\sqrt{(3+3{)}^{2}+(4+4{)}^{2}}=\sqrt{{6}^{2}+{8}^{2}}=\sqrt{36+64}=\sqrt{100}=10$

For a point moving on a circle, the maximum distance from an external point is

$VC+r$

So,

$=10+2=12$

Therefore, the correct answer is

$12$

So, Option C is correct.

Solutions · 1