Question

The equation of circumcircle of the triangle formed by the lines $x=0,y=0,2x+3y=5$, is

• A. $6(x^2+y^2)+5(3x–2y)=0$
• B. $x^2+y^2–2x–3y+5=0$
• C. $x^2+y^2+2x–3y–5=0$
• D. $6(x^2+y^2)–5(3x+2y)=0$

Answer 1

The correct option is D

$6(x^2+y^2)–5(3x+2y)=0$

Explanation for the correct option:

Finding equation of circumcircle of the triangle

The given lines which form the triangle

$$\begin{gathered} x=0.....\left(i\right) \\ y=0....\left(ii\right) \\ 2x+3y=5....\left(iii\right) \end{gathered}$$

From $\left(i\right)\&\left(iii\right)$ we have $\left(0,\frac{5}{3}\right)$

And from $\left(ii\right)\&\left(iii\right)$ we have $\left(\frac{5}{2},0\right)$

So the given triangle has vertices $(0,0)$, $\left(0,\frac{5}{3}\right)$ and $\left(\frac{5}{2},0\right)$

Considering $(x,y)$ as center of the circumcircle

Then, distance of $(x,y)$ from $(0,0)$

$D_1=\sqrt{x^2+y^2}$

The, distance of $(x,y)$ from $\left(0,\frac{5}{3}\right)$

$D_2=\sqrt{x^2-{\left(y-\frac{5}{3}\right)}^2}$

The, distance of $(x,y)$ from $\left(\frac{5}{2},0\right)$

$D_3=\sqrt{{\left(x-\frac{5}{2}\right)}^2+y^2}$

We know for circumcircle $D_1=D_2=D_3$

If $D_1=D_2$

$$\begin{gathered} \Rightarrow \sqrt{x^2+y^2}=\sqrt{x^2+{\left(y-\frac{5}{3}\right)}^2} \\ \Rightarrow x^2+y^2=x^2+y^2+\frac{25}{9}-\frac{10}{3}y[squaringbothsides] \\ \Rightarrow \frac{25}{9}-\frac{10}{3}y=0 \\ \Rightarrow y=\frac{5}{6}...\left(iv\right) \end{gathered}$$

If $D_1=D_3$

$$\begin{gathered} \Rightarrow \sqrt{x^2+y^2}=\sqrt{{\left(x-\frac{5}{2}\right)}^2+y^2} \\ \Rightarrow x^2+y^2=x^2+\frac{25}{4}-5x+y^2[\mathbf{squaring}\mathbf{}\mathbf{both}\mathbf{}\mathbf{sides}] \\ \Rightarrow \frac{25}{4}-5x=0 \\ \Rightarrow x=\frac{5}{4}...\left(v\right) \end{gathered}$$

Thus, the equation has circumcenter $\left(\frac{5}{4},\frac{5}{6}\right)$

Hence its radius is distance from origin of the point $\left(\frac{5}{4},\frac{5}{6}\right)$

$=\sqrt{\frac{25}{16}+\frac{25}{36}}$

Thus, the equation of circle be

${\left(x-\frac{5}{4}\right)}^2+{\left(y-\frac{5}{6}\right)}^2={\left(\sqrt{\frac{25}{16}+\frac{25}{36}}\right)}^2$

$\Rightarrow$$6(x^2+y^2)–5(3x+2y)=0$

Hence, option D is the correct answer.