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If a point P has co-ordinates (0,-2) and Q is any point on the circle, x 2 + y 2 -5x-y+5=0, then the maximum value of (PQ) 2 is:

If a point P has co-ordinates (0,-2) and Q is any point on the circle, x2+ y2-5x-y+5=0, then the maximum value of (PQ)2 is:

Grade:12th pass

3 Answers

Nitesh
38 Points
5 years ago
P(0,-2) and Q on circle x²+y²-5x-y+5=0So centre of circle is (5/2 , 1/2)Which can directly find by trick ( general equation of circle is ax²+by²+2gx+2fy+c=0 so centre is (-g/2,-f/2))So farthest point is always perpendicular line so slope of line with point (5/2 , 1/2) and (0,-2) is -1 .So its perpendicular line slope is 1 .So equation of line which is farthest from circle is ( y-0)=1(x+2)= x-y+2 ans
Nitesh
38 Points
5 years ago
P(0,-2) and Q on circle x²+y²-5x-y+5=0So centre of circle is (5/2 , 1/2)Which can directly find by trick ( general equation of circle is ax²+by²+2gx+2fy+c=0 so centre is (-g/2,-f/2) and radius is (g²+f²-c))So radius is (3/2)½ and total length from center to (0,-2) is. 5/(2)½. So distance PQ is (5 - (3)½)/ (2)½ and max value of (PQ)² is {14 - 5(3)½}.ans
Soumyadeep
15 Points
4 years ago
as per the given equation of circle x^2+y^2-5x-y+5=O and the given point be P(0,-2);
we know that centre of the circle is (-g,-f) from general equation of circle
now on comparing both we get centre coordinate like (5/2,1/2); 
we know that radius of the circle is =√(g²+f²-c) from general equation
therefore radius of this circle is (3½/2½)
distance of the point P from centre is (5/2½)
Now For PQ to be maximum the distance should be equal = the radius + the distance of the point from centre)
PQ =(5+3½)/2½
PQ²=14+5√3

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