Let the equation of the circle, which touches x -axis at the point (a, 0), a 0 and cuts off an intercept of length b on y -axis be x^2+y^2- x+ y+ =0 . If the circle lies below x -axis, then the ordered pair (2 a, b^2 ) is equal to

Step-by-step solution

By pythagoras r ^2= a ^2+ b ^2 4 = P ^2r= 4 a^2+b^2 4 Equation of circle is (x- )^2+(y- )^2=r^2x^2+y^2-2 a x-2 p y+ ^2+p^2-r^2=0 comparision x^2+y^2- x+ y+r=0 array r - =-2 a, =-2 p, r=a^2 2 a= , 4 a^2+b^2=4 p^2 ^2+b^2=4 p^2 ^2+b^2= ^2 array So, (2 a , b ^2 )= ( , ^2-4 r )