Equation of a circle tangent to two lines , given the radius .

Question

What is the equation of the circle whose center is in the first quadrant and with the radius of $4$ units, given that it is tangent to the $x$-axis and to the line $4x-3y=0$?

Answer 1

Do you know the equation for a circle with center $(a,b)$ and radius $4$? You are being asked to evaluate $a,b$. You are given two properties of the circle to do so. The first one, tangent to the $x$ axis, gives you $b$ The other, you need to solve the equation of the circle simultaneously with the equation of the line. For most values of $a$ you will get zero or two solutions, but for one value you will get a single one.

Answer 2

Hint:

Since the radius is $r=4$ and the circle touch the$x$ axis, its center is a point with coordinates $C=(\alpha,4)$. So its equation is: $$ (x-\alpha)^2+(y-4)^2=16 $$ with $\alpha>0$ to be in the first quadrant.

You want that it is tangent to the line $4x-3y=0$ and this means that the system: $$ \begin{cases} (x-\alpha)^2+(y-4)^2=16\\ 4x-3y=0 \end{cases} $$ has only one solution. So, take $x$ (or $y$) from the second equation, substitute in the first and take the discriminant $\Delta$ of the second degree equation that you find. To have a single solution it must be $\Delta=0$. Solve the equation ( that is an equation in $\alpha$) and you have the center of the circle.

Answer 3

Note that the line $y = \frac{4x}{3} -\frac{20}{3} $ is at a distance of 4 units from the line $y=\frac{4x}{3}$ and is parallel to it. Similarly the line $y=4$ is parallel to x axis and is at a distance of 4 units from.

The point where these 2 lines meet will give you the center of the circle. It will be at a distance of 4 units from both the lines which makes sense.

Solving the 2 equations will give you the center as (8,4).

Hence the circle is, $$(x-8)^2+(y-4)^2=16$$

Answer 4

You can do some geometry first to simplify computations: if the $x$-axis and the line $L$ with equation $4x-3y=0$ are tangent to a circle, the straight line through the origin and the centre of the circle is a bissectrix of the geometric angle determined by the $x$-axis and $L$. Let $\theta$ be the polar angle of this bissectrix, $t=\tan\dfrac\theta2$ its slope.

By the duplication formula for tangents, one has $$\tan 2\theta=\frac43=\frac{2t}{1-t^2}, \enspace\text{whence}\quad 4(1-t^2)=6t\iff 2t^2+3t-2=0.$$ Now this quadratic polynomial factors as $\;(2t-1)(t+2)$, hence the bissectrices have equations $$y=\frac x2,\quad y=-2x\quad\text{respectively}.$$ The centres of the corresponding circles are the points on the lines with ordinate $4$, i.e. the points $(8,4)$ and $(-2,4)$, and the circles have equations $$(x-8)^2+(t-4)^2=16, \quad (x+2)^2+(yy-4)^2=16.$$ The circle in the first quadrant is the first one.