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. $2 x + 2 y = 5$
• D. None of these

Answer 1

Answer: (C)

$2 x + 2 y = 5$

Answer 2

The other end is $( t , 3 - t )$

So the equation of the variable circle is

$( x - 1 ) ( x - t ) + ( y - 1 ) ( y - 3 + t ) = 0$

or $x ^ { 2 } + y ^ { 2 } - ( 1 + t ) x - ( 4 - t ) y + 3 = 0$

∴ The centre $( \alpha , \beta )$ is given by $\alpha = \frac { 1 + t } { 2 } , \beta = \frac { 4 - t } { 2 }$

⇒ $2 \alpha + 2 \beta = 5$

Hence, the locus is $2 x + 2 y = 5$ .