Question

If one end of the diameter is $(1,1)$ and other end lies on the line $x+y=3$, then locus of centre of circle is

• A. x + y = 1
• B. 2(x - y) = 5
• C. 2x + 2y = 5
• D. None of these

Answer 1

The correct option is C 2x + 2y = 5

Given one end of the diameter is $(1,1)$and the other end is $x+y=3$
$\Rightarrow y=3-x$
Thus, the parametric form of other end is ($t,3-t$), when $x=t$
So the equation of the variable circle is
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$; where $(x_1,y_1)=(1,1)$ and $(x_2,y_2)=(t,3-t)$
$\Rightarrow (x-1)(x-t)+(y-1)(y-3+t)=0$
$\Rightarrow x^2-tx-x+t+y^2-y(3-t)-y+3-t$
or $x^2+y^2-(1+t)x-(4-t)y+3=0$
$\therefore$ The centre $(\alpha ,\beta )$ is given by
$\alpha =\frac{1+t}{2},\beta =\frac{4-t}{2}$
$\Rightarrow 2\alpha =1+t,2\beta =4-t$
Adding $2\alpha$ and $2\beta$, we get
$\Rightarrow 2\alpha +2\beta =1+t+4-t$
$\Rightarrow 2\alpha +2\beta =5$
Hence, the locus is $2x+2y=5$.