Q. If one end of the diameter is (1, 1) and the 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
Explanation
(c)
Let the other end be $(t,\ 3 - t)$
So, the equation of the variable circle is
$(x - 1)(x - t) + (y - 1)(y - 3 + t) = 0$
$\Rightarrow \ \ \ x^{2} + y^{2} - (1 + t)x - (4 - t)y + 3 = 0$
$\therefore$ The centre $(\alpha,\ \beta)$ is given by
$\alpha = \frac{1 + t}{2},\beta = \frac{4 - t}{2}\ \ \Rightarrow \ \ \ 2\alpha + 2\beta = 5$
Hence, the locus is $2x + 2y = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.