Kabegamimpzo

Expand the equation of the circle #x^2 2x 1 y^2 2y 1 = 25# #x^2 y^2 2x 2y 2 = 25# Differentiate both sides with respect to x using implicit differentiation and the power rule #d/dx(x^2 y^2 2x 2y 2) = d/dx(25)# #2x 2y(dy/dx) 2 2(dy/dx) = 0# #2y(dy/dx) 2(dy/dx) = 2 2x# #dy/dx(2y 2) = 2 2x# #dy/dx = (2 2x)/(2y 2)#Let coordinates of that point be (x,y) \cfrac{y8}{x4} \cfrac{y0}{x2} = 1 And use the fact that (x,y) lies on the circle \implies (x4)^2 (y8)^2 = Solve for x and y Identifying the quadrant in which the parabola lies (or finding the vertex of parabola)√ 25 − (x−4)2 So when we plot these two equations we should have a circle Ifa X Y X2 Y2 25 Andb X Y X2 9y2 Y2 144 Thena Cap B Containsa One Pointb Four Pointsc Two Pointsd Nonecorrect Answer Is Option B Can You Explain This Answer Edurev Jee Question Let the circle (x-1)^2+(y-2)^2=25