Question Statement

A circle touches the line $2x-3y+1=0$ at the point $(1,1)$ and passes through the point of intersection of the lines $x+y+1=0$ and $x-3y+5=0$. Find the equation of the circle.

---

Background and Explanation

To solve this problem, we use the property that the center of a circle $(h,k)$ is equidistant from any two points on its circumference. Additionally, the radius drawn to the point of tangency is always perpendicular to the tangent line. We also utilize the method of solving simultaneous linear equations to find points of intersection.

---

Solution

Let $C(h,k)$ be the center of the required circle. Let $B(1,1)$ be the point where the circle touches the given line, and let $A$ be the point of intersection of the two given lines.

Step 1: Find the point of intersection $A$

We are given the lines: $x+y+1=0\text{— (i)}$ $x-3y+5=0\text{— (ii)}$

Subtracting equation (ii) from equation (i): $\begin{array}{rl}(x+y+1)-(x-3y+5)& =0\\ 4y-4& =0\\ y& =1\end{array}$

Substitute $y=1$ into equation (i): $\begin{array}{rl}x+1+1& =0\\ x+2& =0\\ x& =-2\end{array}$ Thus, the point of intersection is $A(-2,1)$.

Step 2: Use the property that $|AC|=|BC|$

Since both $A(-2,1)$ and $B(1,1)$ lie on the circle, the distance from the center $C(h,k)$ to these points must be equal (both equal the radius $r$). $|AC|=|BC|$ $\sqrt{(h-(-2){)}^2+(k-1{)}^2}=\sqrt{(h-1{)}^2+(k-1{)}^2}$

Squaring both sides: $\begin{array}{rl}(h+2{)}^2+(k-1{)}^2& =(h-1{)}^2+(k-1{)}^2\\ (h+2{)}^2& =(h-1{)}^2\\ h^2+4h+4& =h^2-2h+1\\ 6h& =-3\\ h& =-\frac{1}{2}\end{array}$

Step 3: Use the tangency property

The line $2x-3y+1=0$ touches the circle at $B(1,1)$. The slope of this tangent line ($m_1$) is found by rearranging to $y=mx+c$: $3y=2x+1⟹y=\frac{2}{3}x+\frac{1}{3}⟹m_1=\frac{2}{3}$

The radius $\overline{BC}$ is perpendicular to this tangent line. Therefore: $(\text{Slope of }\overline{BC})\cdot (\text{Slope of tangent})=-1$ $\begin{array}{rl}\left(\frac{k-1}{h-1}\right)\left(\frac{2}{3}\right)& =-1\\ 2(k-1)& =-3(h-1)\\ 2k-2& =-3h+3\\ 3h+2k& =5\end{array}$

Substitute the value of $h=-\frac{1}{2}$: $\begin{array}{rl}3\left(-\frac{1}{2}\right)+2k& =5\\ -\frac{3}{2}+2k& =5\\ 2k& =5+\frac{3}{2}\\ 2k& =\frac{13}{2}\\ k& =\frac{13}{4}\end{array}$ The center of the circle is $C\left(-\frac{1}{2},\frac{13}{4}\right)$.

Step 4: Find the radius squared ($r^2$)

Using point $B(1,1)$ and center $C\left(-\frac{1}{2},\frac{13}{4}\right)$: $\begin{array}{rl}{r}^2& ={\left(1-\left(-\frac{1}{2}\right)\right)}^2+{\left(1-\frac{13}{4}\right)}^2\\ & ={\left(\frac{3}{2}\right)}^2+{\left(-\frac{9}{4}\right)}^2\\ & =\frac{9}{4}+\frac{81}{16}\\ & =\frac{36+81}{16}=\frac{117}{16}\end{array}$

Step 5: Find the equation of the circle

Using the standard form $(x-h{)}^2+(y-k{)}^2=r^2$: $\begin{array}{rl}{\left(x-\left(-\frac{1}{2}\right)\right)}^2+{\left(y-\frac{13}{4}\right)}^2& =\frac{117}{16}\\ {\left(x+\frac{1}{2}\right)}^2+{\left(y-\frac{13}{4}\right)}^2& =\frac{117}{16}\\ x^2+x+\frac{1}{4}+y^2-\frac{13}{2}y+\frac{169}{16}& =\frac{117}{16}\\ x^2+y^2+x-\frac{13}{2}y+\frac{4}{16}+\frac{169}{16}-\frac{117}{16}& =0\\ x^2+y^2+x-\frac{13}{2}y+\frac{56}{16}& =0\\ x^2+y^2+x-\frac{13}{2}y+\frac{7}{2}& =0\end{array}$

Multiply the entire equation by 2 to clear the denominators: $2x^2+2y^2+2x-13y+7=0$

---

Key Formulas or Methods Used

• Point of Intersection: Solved by simultaneous equations.
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Perpendicular Slopes: $m_1\cdot m_2=-1$
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$
• Slope of a Line: For $ax+by+c=0$, slope $m=-\frac{a}{b}$

---

Summary of Steps

1. Find the intersection point $A(-2,1)$ of the two given lines.
2. Set the distance from the center $C(h,k)$ to point $A$ equal to the distance from $C$ to the tangency point $B(1,1)$ to find $h=-1/2$.
3. Use the perpendicularity condition between the radius $BC$ and the tangent line to find the value of $k=13/4$.
4. Calculate the radius squared ($r^2$) using the coordinates of the center and point $B$.
5. Substitute $h,k,$ and $r^2$ into the standard circle equation and simplify to the general form.