Question

Find the equation of the circle whose centre is (3, -2) and which cuts off an intercept of length 6 on the line 4x - 3y + 2 =0 ​

Answer 1

x 2 −6x+y 2 +4y−12=0 (General form).

\large\underline{\text{Main idea}}

Main idea

The radius is a perpendicular bisector to the chord. (Picture attached.)

\large\underline{\text{Explanation}}

Explanation

The pieces of information about the circle are,

\begin{gathered}\boxed{\begin{aligned}&1\text{. Its center is (3,-2).}\\\\&2\text{. $4x-3y+2=0$ cuts through the circle to form a chord of length 6.}\end{aligned}}\end{gathered}

1. Its center is (3,-2).

2. 4x−3y+2=0 cuts through the circle to form a chord of length 6.

We should obtain the distance of (3,-2)(3,−2) to the line 4x-3y+2=04x−3y+2=0 from the distance formula,

\cdots \longrightarrow \boxed{d=\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}.}⋯⟶

d= a 2 +b 2∣ax 1 +by 1 +c∣

Now we get,

\cdots \longrightarrow d=\dfrac{|12+6+2|}{\sqrt{4^{2}+3^{2}}}⋯⟶d= 4 +32

∣12+6+2∣

\cdots \longrightarrow d=\dfrac{20}{5}⋯⟶d= 520

\cdots \longrightarrow d=4.⋯⟶d=4.

There forms a right triangle that consists of the equally divided chord, radius, and perpendicular line.

Let's solve for the radius r.r.

\cdots \longrightarrow r^{2}=3^{2}+4^{2}⋯⟶r

2 =3 2 +4 2

\cdots \longrightarrow r=5.⋯⟶r=5.

Hence,

\cdots \longrightarrow \boxed{r=5.}⋯⟶ r=5.

And hence, the equation of the circle is,

\cdots \longrightarrow \boxed{(x-3)^{2}+(y+2)^{2}=25.}⋯⟶

Answer 2

Find the equation of the circle whose centre is (3, -2) and which cuts off an intercept of length 6 on the line 4x - 3y + 2 =0