Question

If a point `P` has coordinates `(0,-2)` and `Q` is any point on the circle, `x^2+y^2 -5x-y+5=0`, then what is the maximum value of `(PQ)^2`?

Answer 1

Answer is (2).

Explanation:

The center of the circle `x^2+y^2-5x-y+5=0` is `(5/2,1/2)` and radius is `sqrt((5/2)^2+(1/2)^2-5)=sqrt(3/2)=sqrt6/2`

Let us put the values of the corrdinates of `P(0,-2)` in the LHS of the equation of circle `x^2+y^2-5x-y+5=0` and we get

`0+4-0+10+5=19` and as it is is greater than `0`, the pint is outside the circle and maximum distance of `P` from any pont on the circle would be,

distance of `P` from centrer of circle plus radius.

Distance of `P(0,-2)` from center at `(5/2,1/2)` is `sqrt((5/2-0)^2+(1/2-(-2))^2)=sqrt(25/4+25/4)=sqrt(50/4)=(5sqrt2)/2`

and hence the maximum value of `(PQ)^2` is

`((5sqrt2)/2+sqrt6/2)^2`

= `(5sqrt2+sqrt6)^2/4`

= `(50+6+20sqrt3)/4=14+5sqrt3`

Hence answer is (2).

graph{(x^2+(y+2)^2-0.01)(x^2+y^2-5x-y+5)=0 [-3.917, 6.083, -2.48, 2.52]}