Equation of a circle given a tanget line, radius and a line that passes through the center

Question

Find the equation of a circle with $~r=\sqrt2~$ tangent to the line $~x+y =3~$ and having its center on the line $~y=4x~$.

Answer 1

Hint Let $(a,4a)$ be the radius.

You need the equations $$\left\{ \begin{array}{lr}(x-a)^2+(y-4a)^2 &=2 \\ x+y&=3 \end{array} \right. $$

To have an unique solution.

Thus, you want the quadratic equation $$(x-a)^2+(3-x-4a)^2=2$$ to have a double solution.

Answer 2

You have the two equations $$\frac{|a+b-3|}{\sqrt{2}}=\sqrt{2}$$ and $$b=4a$$ where $(a,b)$ the coordinates of the middle point of the circle.

Answer 3

Let $C(x_0,4x_0)$ the centre of the circle, I have: $$(x-x_0)^2+(y-4x_0)^2=2$$ Now imposing the tangency to the line $x+y=3$, I obtain: $$\left\{\begin{matrix} (x-x_0)^2+(y-4x_0)^2 &=2 \\x+y=3 \end{matrix}\right.$$

Substituing, I obtain: $$x^2+x_0^2-2xx_0+9+x^2+16x_0^2-6x+8x_0x-24x_0-2=0$$

Imposing the discriminat equal $0$, I have: $$36(x_0-1)^2-8(17x_0^2-24x_0+7)=0$$ So: $$x_0=\frac{1}{5}\lor x_0=1$$ and $$y_0=\frac{4}{5}\lor y_0=4$$