Find equation of a circle through point that touches two given lines

Question

We have given point of a circle $A(0, 0)$ and two lines: $x+y+2=0 \text { and } x-y+4=0$. We should find equation of a circle that passes through the point $A$ and touches those two lines.

I started solving my making system of three quadratic equations with three unknowns , but this system seems to be very hard to be solvable. Here is what I got so far.

Since point $A$ lies on the circle: $(0 - p)^2 + (0 - q)^2 = r^2, p^2 + q^2 = r^2$. We use the condition for touching of line and circle $r^2(A^2 + B^2) = (Ap + Bq + C)^2$.

$$p^2 + q^2 = r^2\\2r^2 = (p+q+2)^2 \\ 2r^2 = (p-q+4)^2 $$

From here: $(p+q+2)^2 = (p-q+4)^2$. I'm stuck to solving this part since canceling the squares gives wrong and only one result.

If we cancel the squares we get $$p+q+2=p-q+4\\2q = 2\\q=1\\r^2 = p^2+1\\2p^2+2=(p+3)^2\\2p^2+2=p^2+6p+9\\p^2-6p-7=0\\p_1 = -1, p_2 = 7$$

Answer 1

Sketch of Solution:

Suppose that $C$ is a circle with center $c$ that touches both lines. Then, the distance from $c$ to both lines is the same. This implies that $c$ lines on an angle bisector of the two lines. (If the two lines are parallel, then the angle bisector bisects the angle at infinity, i.e., is half-way between the two lines). For any point on the angle bisector, there exists a circle that touches both lines.

In your case, you have two lines with slopes $\pm 1$ which intersect at $(-3,1)$. There are two bisectors, the lines $y=1$ and $x=-3$. Let's work with $y=1$. For any point on this line, i.e., $(t,1)$, you need the distance between $(t,1)$ and a point on one of these lines. The closest point on the line $y=x+4$ is the point of intersection of the line $y=x+4$ with the line $y=-x+(t+1)$, i.e., the line passing through the point $(t,1)$ perpendicular to the line $y=x+4$. The intersection point has $x$ coordinate $\frac{t-3}{2}$. From here, the $y$ coordinate is $\frac{t+5}{2}$, and the radius of the circle with center $(t,1)$ is the distance between $(t,1)$ and $\left(\frac{t-3}{2},\frac{t+5}{2}\right)$. Now, do the same thing with the other bisector.

Finally, you need the condition that the circle contains $(0,0)$, i.e., the point $A$. What you need now is that the radius of the circle equals the distance between $(t,1)$ and the origin. This results in a univariate polynomial equation, which must be solved.

Answer 2

Lines $BT_1$ and $BT_2$ cross at the right angle in the point $B(-3,1)$. The center of the circle must be located on the bisector $BO_1$ of the angle $T_1BT_2$.

There are possibly two solutions, with radii $r_1$ and $r_2$.

Consider the coordinates of the center $O_1$ of the big circle and the tangent point $T_1$:

\begin{align} O_1&=(B_x+r\sqrt2,B_y)=(-3+r\sqrt2,1) ,\\ T_1&= (B_x+\tfrac r{\sqrt2},B_y+\tfrac r{\sqrt2}) = (-3+\tfrac r{\sqrt2},1+\tfrac r{\sqrt2}) . \end{align}

\begin{align} |O_1T_1|^2&=|O_1A|^2=r^2 ,\\ r^2-6\sqrt2\,r+10&=0 , \end{align}

\begin{align} r_{1,2}&=3\sqrt2\pm2\sqrt2 ,\\ r_1&=5\sqrt2 ,\\ r_2&=\sqrt2 ,\\ O_1&=(7,1) ,\\ O_2&=(-1,1) . \end{align}

Answer 3

This specific case is much easier than the general case because the lines make both a 45 degree angle with the $x$ axis. The lines intersect at $(-3, 1)$, so that the center of the circle lies on the angle bisector (which is y=1). x=-3 is also an angle bisector, but it doesn't give any solutions as the origin will be too far away from the center of the circle (origin needs to be in the same quadrant that the two lines split the plane into).

Let the center of the circle be located at $(x-3, 1)$. By the Pythagorean Theorem the distance from the lines is $x/\sqrt{2}$.

Then the distance from the origin is the same, so the square of the distance to the origin equals the square of the distance from the two lines.

This gives $(x-3)^2 + 1^2 = x^2 / 2$, which has solutions $x=2$ and $x=10$.

The coordinates of the center is either $(-1, 1)$, or $(7,-1)$. Both of these give valid circles; the first one has radius $\sqrt{2}$ and the latter has radius $\sqrt{50}$. Both pass through the origin and are tangent to both lines.

The first circle, $(x+1)^2 + (y-1)^2 = 2$ is tangent to the two lines at $(-2, 0)$ and $(-2, 2)$. The second circle, $(x-7)^2 +(y-1)^2 = 50$ is tangent to the two lines at $(2, -4)$ and $(2, 6)$.