# finding the equation of a circle which touches the circle $x^2+y^2-2x-5y=0$ at the origin and passes through the point $(2,1)$ [closed]

Find the equation of the circle which touches the circle $x^2+y^2-2x-5y=0$ at the origin and passes through the point $(2,1)$?

The equation of a circle with center at (a, b) and radius r is $(x- a)^2+ (y- b)^2= r^2$. The circle that passes through the origin must satisfy $(0- a)^2+ (0- b)^2= a^2+ b^2= r^2$. The circle that passes through the point (2, 1) must satisfy $(2- a)^2+ (1- b)^2= 4- 4a+ a^2+ 1- 2b+ b^2= a^2+ b^2- 4a- 2b+ 5= r^2$. That gives two equation for the three values of a, b, and r. The third equation can be derived from the condition that the two circles "touch" which I take to mean "are tangent" at (0 0). The circle $x^2+ y^2- 5y= 0$ slope given by $2x+ 2yy'- 4y'= 0$ so $(2y-4)y'= 2x$ and $y'= 2x/(2y- 4)$. y'= 0 at the (0, 0). For the circle $(x- a)^2+ (y- b)^2= r^2$, $2(x- a)+ 2(y- b)y'= 0$ so $y'= -\frac{y- b}{x- a}$. At (0, 0) that is $y'= -\frac{b}{a}= 0$ so b= 0.
• You made a mistake at the end: $a=0$ and $b=5/2$, which corresponds to the circle $x^2+y^2-5y=0$ (the given circle). – Math Lover Apr 26 '18 at 19:32