Question

The lines $2x–3y=5$ and $3x–4y=7$ are diameters of a circle of area $154sq.units$. Then the equation of this circle is

• A. $x^2+y^2+2x–2y=62$
• B. $x^2+y^2+2x–2y=47$
• C. $x^2+y^2–2x+2y=47$
• D. $x^2+y^2–2x+2y=62$

Answer 1

The correct option is C

$x^2+y^2–2x+2y=47$

Find the equation of the circle based on the given information

Given, equations of diameters are,

$$\begin{gathered} 2x–3y=5....\left(1\right) \\ 3x-4y=7....\left(2\right) \end{gathered}$$

The intersection of two diameters is the centre of the circle

So, the coordinates of the centre will be

$$\begin{gathered} 2x-3y=5 \\ \Rightarrow x=\frac{5+3y}{2} \end{gathered}$$

Putting the above value in $2$

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

Coordinates of the centre is $(1,-1)$

Now, the radius of the circle is,

$$\begin{gathered} \Rightarrow {\mathrm{\pi r}}^2=154 \\ \Rightarrow \mathrm{r}=7 \end{gathered}$$

The equation of the circle $={(x-1)}^2+{(y+1)}^2=7^2$

$\Rightarrow x^2+y^2-2x+2y=47$

Hence, the correct option is $C$, $\Rightarrow x^2+y^2-2x+2y=47$.