Question

If the lines $2x-3y=5$ and $3x-4y=7$ are two diameters of a circle of radius $7$, then the equation of the circle is:

• A. $x^2+y^2+2x-4y-47=0$
• B. $x^2+y^2=49$
• C. $x^2+y^2-2x+2y-47=0$
• D. $x^2+y^2=17$

Answer 1

The correct option is C

$x^2+y^2-2x+2y-47=0$

Explanation for the correct option

Step 1: Find the center of the circle based on given information

Let center of the circle be $(x_0,y_0)$

The two lines given are $2x-3y=5....\left(1\right)$ and $3x-4y=7$

Replace $x=\frac{5+3y}{2}$ from equation $\left(1\right)$ in $3x-4y=7$, we get,

$$\begin{gathered} \Rightarrow 3\left(\frac{5+3y}{2}\right)-4y=7 \\ \Rightarrow \frac{15}{2}+\frac{9y}{2}-4y=7 \\ \Rightarrow \frac{y}{2}=\frac{-1}{2} \\ \Rightarrow y=-1 \end{gathered}$$

Thus, $x=\frac{5+3(-1)}{2}=1$

As the two lines form the diameter of the circle, so $x$ and $y$ will be the center.

So, $\left(x_0,y_0\right)=\left(x,y\right)=\left(1,-1\right)$

Step 2: Evaluate the equation of the circle

The equation of a circle is given by, ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=r^2$, where $(x_0,y_0)$ is the center of the circle.

$$\begin{gathered} \Rightarrow {\left(x-1\right)}^2+{\left(y+1\right)}^2=7^2\left[\because \mathrm{r}=7\right] \\ \Rightarrow x^2+1-2x+y^2+1+2y=49 \\ \Rightarrow x^2+y^2-2x+2y=47 \end{gathered}$$

Therefore, option (C) i.e. $x^2+y^2-2x+2y-47=0$ is the correct answer.