Find the equation of the circle which touches the circle $ x^2 + y^2 = 4 $ and the line $ x+ y =\frac{5}{\sqrt{2}}$

Question

Find the equation of the circle which touches the circle $$ x^2 + y^2 = 4 $$ and the line $$ x+ y =\frac{5}{\sqrt{2}}.$$ I've tried to solve the question by taking the general equation of a circle and eliminate the variables using the data given but, I end up with more variables than equations.

Answer 1

As @IvanAbraham has pointed out, there are an infinite number of possible solutions.

However, there is exactly one such solution where the circle lies exactly in-between the original circle and the line.

See this plot: https://www.desmos.com/calculator/egn1oiv5m9

To get its equation, it's a matter of simple geometry.

Note first that the line of identity $y=x$ will pass through the centre of this little circle, as well as the origin (centre of the large circle) and also be perpendicular to the line.

You can work out the points of intersection of the line of identity with the original large circle as $(\sqrt 2, \sqrt 2)$ (point $P_1$) and with the original line as $(\frac{5\sqrt 2}{4}, \frac{5\sqrt 2}{4})$ (point $P_2$).

And from this, it's easy to deduce that the centre of the little circle will be the midpoint of $P_1$ and $P_2$ and that its diameter will be the distance between $P_1$ and $P_2$. THe radius can be immediately calculated by halving the diameter.

That should give you the final equation as:

$$(x-\frac{9\sqrt 2}{8})^2 + (y-\frac{9\sqrt 2}{8})^2 = \frac{1}{16}$$

Like I said, this is just one of the infinite number of possible solutions that satisfies your problem statement, but it's "special" in a sense. This little circle has the minimum diameter satisfying the constraint.