Question Statement

The diagram shows a rectangle $PQRS$ and the circles $C_1$ and $C_2$. Both the circles touch each other and three sides of the rectangle. The coordinates of the points $P,Q,R$ and $S$ are $(0,4)$, $(1,1)$, $(7,3)$ and $(6,6)$. Find the equation of the circles $C_1$ and $C_2$.

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Background and Explanation

To solve this problem, we use the properties of a rectangle and tangent circles. We need the Midpoint Formula to find the center line of the rectangle, the Section (Ratio) Formula to locate the centers of the circles along that line, and the Distance Formula to determine the radius. Finally, we apply the Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$.

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Solution

Let $M$ and $N$ be the midpoints of the sides $\overline{PQ}$ and $\overline{RS}$ of the rectangle, respectively. Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$.

Step 1: Find the midpoints of the rectangle's widths

The midpoints $M$ and $N$ define the central axis of the rectangle upon which the centers of the circles must lie. $M=\left(\frac{0+1}{2},\frac{4+1}{2}\right)=\left(\frac{1}{2},\frac{5}{2}\right)$ $N=\left(\frac{6+7}{2},\frac{6+3}{2}\right)=\left(\frac{13}{2},\frac{9}{2}\right)$

Step 2: Determine the radius of the circles

Since the circles touch three sides of the rectangle, the diameter of each circle is equal to the length of the side $PQ$. The radius $r$ is half the distance of $PQ$: $\begin{array}{rl}r=\frac{1}{2}|PQ|& =\frac{1}{2}\sqrt{(0-1{)}^2+(4-1{)}^2}\\ & =\frac{1}{2}\sqrt{1+9}\\ & =\frac{1}{2}\sqrt{10}\end{array}$

Step 3: Find the coordinates of the centers $A$ and $B$

The total length $MN$ is equivalent to two diameters (or 4 radii). The center $A$ is one radius away from $M$, and center $B$ is three radii away from $M$. Therefore, $A$ divides the line segment $\overline{MN}$ in the ratio $1:3$. Using the ratio formula: $\begin{array}{rl}A& =\left(\frac{1\left(\frac{13}{2}\right)+3\left(\frac{1}{2}\right)}{1+3},\frac{1\left(\frac{9}{2}\right)+3\left(\frac{5}{2}\right)}{1+3}\right)\\ & =\left(\frac{\frac{13}{2}+\frac{3}{2}}{4},\frac{\frac{9}{2}+\frac{15}{2}}{4}\right)\\ & =\left(\frac{8}{4},\frac{12}{4}\right)=(2,3)\end{array}$

Similarly, $B$ divides the line segment $\overline{MN}$ in the ratio $3:1$: $\begin{array}{rl}B& =\left(\frac{3\left(\frac{13}{2}\right)+1\left(\frac{1}{2}\right)}{3+1},\frac{3\left(\frac{9}{2}\right)+1\left(\frac{5}{2}\right)}{3+1}\right)\\ & =\left(\frac{\frac{39}{2}+\frac{1}{2}}{4},\frac{\frac{27}{2}+\frac{5}{2}}{4}\right)\\ & =\left(\frac{20}{4},\frac{16}{4}\right)=(5,4)\end{array}$

Step 4: Find the equation of circle $C_1$

Using center $A(2,3)$ and radius $r=\frac{1}{2}\sqrt{10}$: $\begin{array}{rl}(x-2{)}^2+(y-3{)}^2& ={\left(\frac{1}{2}\sqrt{10}\right)}^2\\ x^2-4x+4+y^2-6y+9& =\frac{10}{4}=\frac{5}{2}\\ 2(x^2+y^2-4x-6y+13)& =5\\ 2x^2+2y^2-8x-12y+26& =5\\ 2x^2+2y^2-8x-12y+21& =0\end{array}$

Step 5: Find the equation of circle $C_2$

Using center $B(5,4)$ and radius $r=\frac{1}{2}\sqrt{10}$: $\begin{array}{rl}(x-5{)}^2+(y-4{)}^2& ={\left(\frac{1}{2}\sqrt{10}\right)}^2\\ x^2-10x+25+y^2-8y+16& =\frac{5}{2}\\ 2(x^2+y^2-10x-8y+41)& =5\\ 2x^2+2y^2-20x-16y+82& =5\\ 2x^2+2y^2-20x-16y+77& =0\end{array}$

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Key Formulas or Methods Used

• Midpoint Formula: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Section (Ratio) Formula: $P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$

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Summary of Steps

1. Calculate the midpoints $M$ and $N$ of the rectangle's shorter sides to find the axis of the centers.
2. Calculate the radius by finding half the length of side $PQ$.
3. Use the ratio formula to find centers $A$ and $B$ along the segment $MN$ (ratios $1:3$ and $3:1$).
4. Substitute the centers and the squared radius into the circle equation.
5. Expand and simplify to the general form $A{x}^2+B{y}^2+Cx+Dy+E=0$.