Question Statement

The two points $A(4,3)$ and $B(2,5)$ lie on a circle. The centre of the circle lies on the perpendicular bisector of the chord $\overline{AB}$. The distance between the centre and the chord $\overline{AB}$ is $\sqrt{7}$. Find the equation of the circle.

---

Background and Explanation

To find the equation of a circle, we need to determine its centre $(h,k)$ and its radius $r$. This problem utilizes the geometric property that the perpendicular bisector of any chord passes through the centre of the circle and uses the distance formula to relate the centre to the given points.

---

Solution

Let $C(h,k)$ be the centre of the circle and let $M$ be the midpoint of the chord $\overline{AB}$.

1. Find the Midpoint $M$ of Chord $\overline{AB}$

Using the midpoint formula for $A(4,3)$ and $B(2,5)$: $M=\left(\frac{4+2}{2},\frac{3+5}{2}\right)=(3,4)$

2. Establish the Relationship between $h$ and $k$

Since the centre $C$ lies on the perpendicular bisector of $\overline{AB}$, the line segment $\overline{CM}$ is perpendicular to the chord $\overline{AB}$. The product of their slopes must be $-1$: $(\text{Slope of }\overline{CM})(\text{Slope of }\overline{AB})=-1$

Calculating the slopes: $\left(\frac{k-4}{h-3}\right)\left(\frac{5-3}{2-4}\right)=-1$ $\left(\frac{k-4}{h-3}\right)\left(\frac{2}{-2}\right)=-1$ $\left(\frac{k-4}{h-3}\right)(-1)=-1$ $k-4=h-3$ $h-k=-1\dots \text{(i)}$

3. Use the Distance from Centre to Chord

The distance between the centre $C(h,k)$ and the midpoint $M(3,4)$ is given as $\sqrt{7}$: $|CM|=\sqrt{7}$ $\sqrt{(h-3{)}^2+(k-4{)}^2}=\sqrt{7}$

Squaring both sides: $(h-3{)}^2+(k-4{)}^2=7$ $h^2-6h+9+k^2-8k+16=7$ $h^2+k^2-6h-8k+18=0\dots \text{(ii)}$

4. Solve for $h$ and $k$

From equation (i), we have $h=k-1$. Substitute this into equation (ii): $(k-1{)}^2+k^2-6(k-1)-8k+18=0$ $k^2-2k+1+k^2-6k+6-8k+18=0$ $2k^2-16k+25=0$

Using the quadratic formula $k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $k=\frac{-(-16)\pm \sqrt{(-16{)}^2-4(2)(25)}}{2(2)}$ $k=\frac{16\pm \sqrt{256-200}}{4}=\frac{16\pm \sqrt{56}}{4}$ $k=\frac{16\pm 2\sqrt{14}}{4}=\frac{8\pm \sqrt{14}}{2}$

This gives two possible values for $k$: $k_1=\frac{8-\sqrt{14}}{2}$ and $k_2=\frac{8+\sqrt{14}}{2}$

Now, find the corresponding values for $h$ using $h=k-1$: For $k_1$: $h_1=\frac{8-\sqrt{14}}{2}-1=\frac{8-\sqrt{14}-2}{2}=\frac{6-\sqrt{14}}{2}$ For $k_2$: $h_2=\frac{8+\sqrt{14}}{2}-1=\frac{8+\sqrt{14}-2}{2}=\frac{6+\sqrt{14}}{2}$

5. Calculate the Radius $r$

The radius is the distance from the centre to point $A(4,3)$.

Case I: Centre is $\left(\frac{6-\sqrt{14}}{2},\frac{8-\sqrt{14}}{2}\right)$ $r=\sqrt{{\left(4-\frac{6-\sqrt{14}}{2}\right)}^2+{\left(3-\frac{8-\sqrt{14}}{2}\right)}^2}$ $r=\sqrt{{\left(\frac{8-6+\sqrt{14}}{2}\right)}^2+{\left(\frac{6-8+\sqrt{14}}{2}\right)}^2}$ $r=\sqrt{{\left(\frac{2+\sqrt{14}}{2}\right)}^2+{\left(\frac{-2+\sqrt{14}}{2}\right)}^2}$ $r=\sqrt{\frac{4+4\sqrt{14}+14}{4}+\frac{4-4\sqrt{14}+14}{4}}=\sqrt{\frac{36}{4}}=\sqrt{9}=3$

Case II: Centre is $\left(\frac{6+\sqrt{14}}{2},\frac{8+\sqrt{14}}{2}\right)$ $r=\sqrt{{\left(4-\frac{6+\sqrt{14}}{2}\right)}^2+{\left(3-\frac{8+\sqrt{14}}{2}\right)}^2}$ $r=\sqrt{{\left(\frac{8-6-\sqrt{14}}{2}\right)}^2+{\left(\frac{6-8-\sqrt{14}}{2}\right)}^2}$ $r=\sqrt{\frac{4-4\sqrt{14}+14+4+4\sqrt{14}+14}{4}}=\sqrt{\frac{36}{4}}=3$

6. Final Equations of the Circle

The standard form is $(x-h{)}^2+(y-k{)}^2=r^2$.

Equation 1: ${\left(x-\frac{6-\sqrt{14}}{2}\right)}^2+{\left(y-\frac{8-\sqrt{14}}{2}\right)}^2=9$

Equation 2: ${\left(x-\frac{6+\sqrt{14}}{2}\right)}^2+{\left(y-\frac{8+\sqrt{14}}{2}\right)}^2=9$

---

Key Formulas or Methods Used

• Midpoint Formula: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$
• Perpendicular Lines: $m_1\cdot m_2=-1$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$

---

Summary of Steps

1. Calculate the midpoint $M$ of the points $A$ and $B$.
2. Determine the slope of $AB$ and find the perpendicular slope to establish a linear equation between $h$ and $k$.
3. Use the given distance $\sqrt{7}$ from the centre to the midpoint to create a second equation.
4. Solve the system of equations to find the two possible coordinates for the centre.
5. Calculate the radius by finding the distance between the centre and one of the points on the circle.
6. Substitute the centre and radius into the standard circle equation.