x^(2)+y^(2)-6x+8y+2=0

If the circle x^(2) +y^(2) -6x -8y+( 25-a^(2) ) =0 touches the axis of x, then a equals.

Find k if the following pairs of circles are orthogonal x^(2)+y^(2)-6x-8y+12=0,x^(2)+y^(2)-8x+6y+k=0

The circles x^(2)+y^(2)-6x-8y=0 and x^(2)+y^(2)-6x+8=0 are

(3,-4) is the point to which the origin is shifted and the transformed equation is X^(2)+Y^(2)=4 then the original equation is (A) x^(2)+y^(2)+6x+8y+21=0 (B) x^(2)+y^(2)+6x+8y-21,=0 (C) x^(2)+y^(2)-6x+8y+21=0 (D) x^(2)+y^(2)-6x-8y+21=0

Through the point P(3,4) a pair of perpendicular lines are drawn which meet x-axis at the point A and B. The locus of incentre of triangle PAB is (a) x^(2)-y^(2)-6x-8y+25=0 (b) x^(2)+y^(2)-6x-8y+25=0 (c) x^(2)-y^(2)+6x+8y+25=0 (d) x^(2)+y^(2)+6x+8y+25=0

If the circles x^(2)+y^(2)=a^(2), x^(2)+y^(2)-6x-8y+9=0 touch externally then a=

If the circles x^(2)+y^(2)=a^(2), x^(2)+y^(2)-6x-8y+9=0 touch externally then a=

If the circle x^(2)+y^(2)+8x-4y+c=0 touches the circle x^(2)+y^(2)+2x+4y-11=0 externally and cuts the circle x^(2)+y^(2)-6x+8y+k=0 orthogonally then k