Suppose that the equation of the circle having `(-3, 5) and (5,-1)` as end points of a diameter is`(x-a)^2+ (y - b)^2 =r^2` . Then `a+b+r`, (r> 0) is: (A) `8 ` (B)`9 ` (C) `10 ` (D) `11`

Find the equation of the circle with centre C and radius r where C = (-a,-b) , r = sqrt(a^(2) -b^(2)) (|a||gt|b|)

Find the equation of the circle, the coordinates of the end points of one of whose diameters are (i) A (3, 2 ) and B(2, 5) (ii) A (5, - 3) and B(2, - 4 ) (iii) A(-2, -3) and B(-3, 5) (iv) A(p, q), B (r, s )

If one end of a diameter of the circle x^2 + y^2 - 8x - 14y+c=0 is the point (-3, 2) , then its other end is the point. (A) (5, 7) (B) (9, 11) (C) (10, 11) (D) (11, 12)

If one end of a diameter of the circle x^2 + y^2 - 8x - 14y+c=0 is the point (-3, 2) , then its other end is the point. (A) (5, 7) (B) (9, 11) (C) (10, 11) (D) (11, 12)

If the two circles (x+1)^2+(y-3)^2=r^2 and x^2+y^2-8x+2y+8=0 intersect at two distinct point,then (A) r > 2 (B) 2 < r < 8 (C) r < 2 (D) r=2

If A D is median of A B C and P is a point on A C such that a r\ ( A D P):\ a r\ ( A B D)=2:3,\ then a r( P D C): a r\ ( A B C) is 1:5 (b) 1:5 (c) 1:6 (d) 3:5

If A D is median of A B C and P is a point on A C such that a r\ ( A D P):\ a r\ ( A B D)=2:3,\ then a r( P D C): a r\ ( A B C) is 1:5 (b) 1:5 (c) 1:6 (d) 3:5