Question

If one end of the diameter is (1,1) and the other end lies on the line $x + y = 3$, then find the locus of the circle.
A. $x + y = 1$
B. $2(x - y) = 5$
C. $2x + 2y = 5$
D. None of these

Answer 1

Hint: First write $x + y = 3$ in terms of y, then obtain a point in this line. Then write the formula of locus of a circle passing through two points. Then obtain the centre of the circle and equate t from two equations.

Formula Used:
Equation of the circle passing through two points $(a,b),(c,d)$ is,
$(x - a)(x - c) + (y - b)(y - d) = 0$ .
Centre of the circle ${x^2} + {y^2} + 2gx + 2fy = c$ is $( - g, - f)$ .

Complete step by step solution:
The given equation of the line is,
$x + y = 3$
$ \Rightarrow y = 3 - x$
The parametric point in this line is $(t,3 - t)$ .
The equation of the circle passing through two points (1,1) and $(t,3 - t)$is,
$(x - 1)(x - t) + (y - 1)(y - 3 + t) = 0$
${x^2} - xt - x + t + {y^2} - 3y + ty - y + 3 - t = 0$
${x^2} + {y^2} - (1 + t)x - (4 - t)y + 3 = 0$
Therefore, the centre of the circle is $\left( {\dfrac{{1 + t}}{2},\dfrac{{4 - t}}{2}} \right)$ .
Suppose the centre is (a, b).
Hence,
$a = \dfrac{{1 + t}}{2}$
$t = 2a - 1$ --(1)
And,
 $b = \dfrac{{4 - t}}{2}$
$t = 4 - 2b$ --(2)
Equate (1) and (2) to obtain the required path.
$2a - 1 = 4 - 2b$
$2a + 2b = 5$
The required locus is,
$2x + 2y = 5$ .

Option ‘C’ is correct

Note: Sometimes we can go with the approach that after having two points, no need to calculate the equation of the circle, by just applying the formula that the centre is the middle point of the diameter. So, we can write $\left( {\dfrac{{1 + t}}{2},\dfrac{{1 + 3 - t}}{2}} \right)$ as the centre. Suppose that centre is (a, b), equate (a, b) with $\left( {\dfrac{{1 + t}}{2},\dfrac{{1 + 3 - t}}{2}} \right)$ to obtain the value of t from both the equation, then equate two equation and obtain the locus.