Is it possible to find intersection of circle and triangle?

Question

Is it possible to find a the exact points a circle will intersect the legs / hypotenuse of a Right 45° triangle?

In the picture above it would be the points that are green which are unknown, but the circle radius and location are known as well as all aspects of the triangle (angles, leg lengths, etc).

Answer 1

Sure. The equation of the circle with center $(a,b)$ and radius $r$ is $$(x-a)^2 + (y-b)^2 = r^2$$ while the equation of the horizontal line is $y = -100$. Plugging it in you get $$(x-73)^2 + (-100+93)^2 = r^2$$ and $r$ is fixed, so there will be $2$ solutions which are your points.

For the hypotenuse, it's the equation $x=-y$ you need, and plugging it in you get another quadratic in $x$...

UPDATE

With $r=19$ you get $$(x-73)^2 = 19^2 - 7^2 = 312$$ which yields $$x =\ 73 \pm \sqrt{312} \approx 73 \pm 17.66 = \{90.66, 55.34 \}$$

For the second one, with $y=-x$ you get $$(x-73)^2 + (-x + 93)^2 = 19^2$$ you get $$x = 83 \pm \sqrt{\frac{161}{2}} \approx \{74.03, 91.97\}.$$

Answer 2

HINT

Note that

• circle equation is $(x-73)^2+(y-93)^2=19^2$
• line equations are $y=100$ and $y=x$
• solve the systems of circle with each line to find intersection points

that is

$$\begin{cases}(x-73)^2+(y-93)^2=19^2\\\\y=100 \end{cases}\implies \begin{cases}(x-73)^2+(100-93)^2=19^2\\\\y=100 \end{cases}$$

and

$$\begin{cases}(x-73)^2+(y-93)^2=19^2\\\\y=x \end{cases}\implies \begin{cases}(x-73)^2+(x-93)^2=19^2\\\\y=100 \end{cases}$$

for each system solve the quadratic equation in $x$ and find the intersection points.

Answer 3

Alternative approach:

Suppose the center of the circle is $I$. Since the coordinates of $I$ is $(73,-93)$, the vertical distance from $I$ to leg $a$ (say $IR$) is: $-93-(-100)=7$.

Suppose two intersection of circle with leg $a$ are $P$ and $Q$, thus $IP=IQ=19$ (since the radius of the circle is $19$). Aware also that it can be proved by geometry of circle that $PR=QR$ and hence, the coordinates of $R$ is $(73,-100)$.

Then, by Pythagorean theorem, $PR^2=PI^2-IR^2$ and $QR^2=QI^2-IR^2$. Thus, $PR^2=19^2-7^2=312$. Therefore, x-coordinate of $P$ and $Q$ are $73-\sqrt{312}$ and $73+\sqrt{312}$, respectively (assume $P$ is closer to y-axis than $Q$).

Therefore, the coordinates of intersections of circle on leg $a$ are $\left((73-\sqrt{312}),-100\right)$ and $\left((73+\sqrt{312}),-100\right)$, respectively. Or approximately, $(55.34,-100)$ and $(90.66,-100)$.

Similarly, suppose two intersection of circle on the hypotenuse of the right-triangle are $S$ and $T$, thus $IS=IT=19$ (the radius of the circle). Since both $S$ and $T$ are on $y=-x$ line, magnitude of the coordinates of each point must be equal but have opposite sign. Let them be $S(m,-m)$ and $T(n,-n)$. Now, suppose the parallel line goes through $I$ meets the perpendicular line goes through $S$ at $U$. Hence,the coordinates of $U$ must be $(m,-93)$. By Pythagorean theorem, $SU^2=SI^2-IU^2$. Therefore, $(m-93)^2=19^2-(73-m)^2$. Thus, $2m^2-(186+146)m+(93^2+73^2-19^2)=0$ or $2m^2-332m+13617=0$. Quadratic equation gives the solutions: $m=74.03$ or $m=91.97$.

Similarly, for point $T(n,-n)$, the Pythagorean theorem would be $TW^2=TI^2-IW^2$ where $W$ is the intersecting point of the line parallel to leg $a$, which goes through $I$ and the perpendicular line goes through $T$. Therefore, $(n-93)^2=19^2-(n-73)^2$, the solution for which would be the same as $n=74.03$ or $n=91.97$.

Thus, the coordinates of intersections of circle on the hypotenuse of the right-triangle are approximately, $(74.03,-74.03)$ (closest to x-axis) and $(91.97,-91.97)$ (farthest to x-axis).