Question Statement

Find the equation of the circle passing through the points $A(1,-2)$ and $B(4,-3)$ and whose centre lies on the line $3x+4y=7$.

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Background and Explanation

To find the equation of a circle, we need its centre $(h,k)$ and its radius $r$. We use the distance formula to equate the radii from the centre to the given points and use the fact that the centre must satisfy the equation of the line it lies on.

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Solution

Let the centre of the circle be $C(h,k)$ and its radius be $r$.

Step 1: Use the condition that the centre lies on the given line

Since the centre $C(h,k)$ lies on the line $3x+4y=7$, it must satisfy the equation: $3h+4k=7$

Step 2: Use the property that distances from the centre to points on the circle are equal

The distance from the centre $C(h,k)$ to point $A(1,-2)$ and point $B(4,-3)$ must be equal to the radius $r$. Therefore, $|AC|=|BC|$, which implies $|AC{|}^2=|BC{|}^2$.

Using the distance formula: $(h-1{)}^2+(k-(-2){)}^2=(h-4{)}^2+(k-(-3){)}^2$ $(h-1{)}^2+(k+2{)}^2=(h-4{)}^2+(k+3{)}^2$

Expanding the squares: $h^2-2h+1+k^2+4k+4=h^2-8h+16+k^2+6k+9$ $-2h+4k+5=-8h+6k+25$

Rearranging the terms to form a linear equation: $6h-2k=20$ Dividing by 2: $3h-k=10$

Step 3: Solve the system of equations for $h$ and $k$

We have the system:

1. $3h+4k=7$
2. $3h-k=10$

Subtracting equation (ii) from equation (i): $(3h+4k)-(3h-k)=7-10$ $5k=-3⟹k=-\frac{3}{5}$

Substitute $k=-\frac{3}{5}$ into equation (ii): $3h-\left(-\frac{3}{5}\right)=10$ $3h+\frac{3}{5}=10$ $3h=10-\frac{3}{5}=\frac{47}{5}$ $h=\frac{47}{15}$

The centre is $C\left(\frac{47}{15},-\frac{3}{5}\right)$.

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

$r^2=|AC{|}^2={\left(\frac{47}{15}-1\right)}^2+{\left(-\frac{3}{5}+2\right)}^2$ $r^2={\left(\frac{32}{15}\right)}^2+{\left(\frac{7}{5}\right)}^2$ $r^2=\frac{1024}{225}+\frac{49}{25}$ To add these, find a common denominator (225): $r^2=\frac{1024}{225}+\frac{441}{225}=\frac{1465}{225}=\frac{293}{45}$

Step 5: Write the equation of the circle

Using the standard form $(x-h{)}^2+(y-k{)}^2=r^2$: ${\left(x-\frac{47}{15}\right)}^2+{\left(y-\left(-\frac{3}{5}\right)\right)}^2=\frac{293}{45}$ ${\left(x-\frac{47}{15}\right)}^2+{\left(y+\frac{3}{5}\right)}^2=\frac{293}{45}$

Expanding the equation: $x^2-\frac{94}{15}x+\frac{2209}{225}+y^2+\frac{6}{5}y+\frac{9}{25}-\frac{293}{45}=0$ $x^2+y^2-\frac{94}{15}x+\frac{6}{5}y+\left(\frac{2209+81-1465}{225}\right)=0$ $x^2+y^2-\frac{94}{15}x+\frac{6}{5}y+\frac{825}{225}=0$ $x^2+y^2-\frac{94}{15}x+\frac{6}{5}y+\frac{11}{3}=0$

Multiply the entire equation by 15 to clear denominators: $15x^2+15y^2-94x+18y+55=0$

This is the required equation of the circle.

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Key Formulas or Methods Used

• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Point-on-Line Property: If a point $(h,k)$ lies on a line $ax+by=c$, then $ah+bk=c$.
• System of Linear Equations: Solving for variables using elimination or substitution.

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Summary of Steps

1. Assign the centre coordinates as $(h,k)$.
2. Substitute $(h,k)$ into the line equation $3x+4y=7$.
3. Set the squared distance from the centre to $(1,-2)$ equal to the squared distance from the centre to $(4,-3)$.
4. Solve the resulting two linear equations to find the values of $h$ and $k$.
5. Calculate $r^2$ using the distance formula with the centre and one of the given points.
6. Substitute $h,k,$ and $r^2$ into the standard circle equation and simplify to general form.