What is the equation of circle with radius $\sqrt{2}$, tangent to the line $x+y=3$, and having its center on the line $y=4x$?

Question

What is the equation of circle with radius $\sqrt{2}$, tangent to the line $x+y=3$, and having its center on the line $y=4x$?

Can someone help me please?

Answer 1

For the circle's center you want a point on the line $\;y=4x\;$, namely of the form $\;(a,4a)\;$ , and a distance of $\;\sqrt2\;$ from the line $\;x+y-3=0\;$ :

$$\frac{|a+4a-3|}{\sqrt{1+1}}=\sqrt2\iff |5a-3|=2\iff\begin{cases}-5a+3=2\iff a=\frac15\\{}\\5a-3=2\iff a=1\end{cases}$$

Complete now.