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?

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1 Answer
Mar 18, 2018

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]}