The center of the circle x^2+y^2-5x-y+5=0x2+y2−5x−y+5=0 is (5/2,1/2)(52,12) and radius is sqrt((5/2)^2+(1/2)^2-5)=sqrt(3/2)=sqrt6/2√(52)2+(12)2−5=√32=√62
Let us put the values of the corrdinates of P(0,-2)P(0,−2) in the LHS of the equation of circle x^2+y^2-5x-y+5=0x2+y2−5x−y+5=0 and we get
0+4-0+10+5=190+4−0+10+5=19 and as it is is greater than 00, the pint is outside the circle and maximum distance of PP from any pont on the circle would be,
distance of PP from centrer of circle plus radius.
Distance of P(0,-2)P(0,−2) from center at (5/2,1/2)(52,12) is sqrt((5/2-0)^2+(1/2-(-2))^2)=sqrt(25/4+25/4)=sqrt(50/4)=(5sqrt2)/2√(52−0)2+(12−(−2))2=√254+254=√504=5√22
and hence the maximum value of (PQ)^2(PQ)2 is
((5sqrt2)/2+sqrt6/2)^2(5√22+√62)2
= (5sqrt2+sqrt6)^2/4(5√2+√6)24
= (50+6+20sqrt3)/4=14+5sqrt350+6+20√34=14+5√3
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]}
