Question Statement

Find the equation of the circle with its centre lying on the line $x+y=2$ and passing through the points $(2,-2)$ and $(3,4)$.

---

Background and Explanation

To find the equation of a circle, we need its centre $(h,k)$ and its radius $r$. We use the property that the distance from the centre to any point on the circle is equal to the radius. If the centre lies on a given line, its coordinates must satisfy the equation of that line.

---

Solution

Let $C(h,k)$ be the centre of the circle.

1. Using the Line Equation

It is given that the centre lies on the line $x+y=2$. Therefore, the coordinates $(h,k)$ must satisfy this equation: $h+k=2$

2. Using the Distance Property

Let the given points be $A(2,-2)$ and $B(3,4)$. Since both points lie on the circle, the distance from the centre to $A$ must equal the distance from the centre to $B$ (both are equal to the radius $r$): $|AC|=|BC|$

Using the distance formula $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$: $\sqrt{(h-2{)}^2+(k+2{)}^2}=\sqrt{(h-3{)}^2+(k-4{)}^2}$

Squaring both sides to remove the square roots: $(h-2{)}^2+(k+2{)}^2=(h-3{)}^2+(k-4{)}^2$

Expanding the squares: $h^2-4h+4+k^2+4k+4=h^2-6h+9+k^2-8k+16$

Cancel $h^2$ and $k^2$ from both sides and simplify: $-4h+4k+8=-6h-8k+25$ $2h+12k=17$

3. Solving for the Centre $(h,k)$

We now have a system of two linear equations:

1. $h+k=2$
2. $2h+12k=17$

Multiply equation (i) by 2: $2h+2k=4$

Subtract this from equation (ii): $(2h+12k)-(2h+2k)=17-4$ $10k=13$ $k=\frac{13}{10}$

Substitute $k=\frac{13}{10}$ back into equation (i): $h+\frac{13}{10}=2$ $h=2-\frac{13}{10}=\frac{20-13}{10}=\frac{7}{10}$

Thus, the centre of the circle is $C\left(\frac{7}{10},\frac{13}{10}\right)$.

4. Finding the Radius

The radius $r$ is the distance $|AC|$: $r=\sqrt{{\left(\frac{7}{10}-2\right)}^2+{\left(\frac{13}{10}+2\right)}^2}$ $r=\sqrt{{\left(-\frac{13}{10}\right)}^2+{\left(\frac{33}{10}\right)}^2}$ $r=\sqrt{\frac{169}{100}+\frac{1089}{100}}=\sqrt{\frac{1258}{100}}$

5. Equation of the Circle

Using the standard form $(x-h{)}^2+(y-k{)}^2=r^2$: ${\left(x-\frac{7}{10}\right)}^2+{\left(y-\frac{13}{10}\right)}^2={\left(\sqrt{\frac{1258}{100}}\right)}^2$

Expanding the terms: $x^2-\frac{7}{5}x+\frac{49}{100}+y^2-\frac{13}{5}y+\frac{169}{100}=\frac{1258}{100}$ $x^2+y^2-\frac{7}{5}x-\frac{13}{5}y+\frac{49+169-1258}{100}=0$ $x^2+y^2-\frac{7}{5}x-\frac{13}{5}y-\frac{1040}{100}=0$ $x^2+y^2-\frac{7}{5}x-\frac{13}{5}y-\frac{52}{5}=0$

Multiply the entire equation by 5 to clear the denominators: $5x^2+5y^2-7x-13y-52=0$

This is the required equation of the circle.

---

Key Formulas or Methods Used

• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Linear Systems: Solving by elimination/substitution.
• Point-on-Line Property: If a point lies on a line, its coordinates satisfy the line's equation.

---

Summary of Steps

1. Define the centre as $(h,k)$ and substitute it into the line equation $x+y=2$.
2. Set the distance from the centre to $(2,-2)$ equal to the distance from the centre to $(3,4)$.
3. Simplify the resulting distance equation to get a second linear equation in terms of $h$ and $k$.
4. Solve the system of two equations to find the coordinates of the centre.
5. Calculate the radius squared ($r^2$) using the distance formula.
6. Substitute the centre and $r^2$ into the standard circle equation and simplify to the general form.