Question

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

• A. $\left(-\frac{5}{2},-\frac{1}{2}\right)$
• B. $\left(\frac{5}{2},-\frac{1}{2}\right)$
• C. $\left(-\frac{1}{2},-\frac{1}{2}\right)$
• D. $\left(\frac{5}{2},\frac{1}{2}\right)$

Answer 1

Answer: (B)

Given equation of sides are
$y=0\dots (i)$
$y=x\dots (ii)$
$2x+3y=10\dots (iii)$

Solving Eqs. (i) and (iii) to get vertex 'A'
$\therefore A=(5,0)$
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 $x^2+y^2+2gx+2fy+c=0\dots (iv)$
Since, Eq. (iv) passes through $B(0,0)$
$\Rightarrow C=0$
Since, Eq. (iv) passes through $A(5,0)$
$25+0+10g+0+0=0$
$g=-5/2$
Eq. (iv) passes through $C(2,2)$
$\therefore 4+4+4g+4y^{\prime \prime }=0$
$g+f+2=0$
$-\frac{5}{2}+f+2=0$
$f=\frac{1}{2}$
$\therefore$ Centre of circle $=(-g,-f)=\left(\frac{5}{2},\frac{-1}{2}\right)$