The centre of the circle which passes through the vertices of the triangle formed by the lines $y=0, y=x$ and $2 x+3 y=10$, is

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Aicharya · Accepted Answer

Given equation of sides are $\begin{aligned} y & =0 \quad \ldots (i) \ y & =x \quad \ldots (ii) \ 2 x+3 y & =10 \quad \ldots (iii) \end{aligned}$ Solving Eqs. (i) and (iii) to get vertex ' $A$ ' $\therefore \quad A=(5,0)$ Solving (i) and (ii) to get vertex ' $B$ ' $B=(0,0)$ Solving Eqs. (ii) and (iii) to get vertex ' $C$ ' $C=(2,2)$ Let equation of circle be $x^2+y^2+2 g x+2 f y+c=0$ ...(iv) Since, Eq. (iv) passes through $B(0,0)$ $\Rightarrow \quad C=0$ Since, Eq. (iv) passes through $A(5,0)$ $\begin{aligned} & 25+0+10 g+0+0=0 \ & g=-5 / 2 \end{aligned}$ Eq. (iv) passes through $C(2,2)$ $\begin{gathered} \therefore \quad 4+4+4 g+4 f=0 \ g+f+2=0 \ -\frac{5}{2}+f+2=0 \ f=\frac{1}{2} \ \therefore \text { Centre of circle }=(-g,-f)=\left(\frac{5}{2}, \frac{-1}{2}\right) \end{gathered}$ $\therefore$ Hence, answer is (b).

The centre of the circle which passes through the vertices of the triangle formed by the lines \(y=0, y=x\)…

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