VerifiedHint: First, we have to draw rough figures just for clear understanding. Then, we have to find the midpoint of the circle with centre O using the formula $\left( a=\dfrac{c+x}{2},b=\dfrac{d+y}{2} \right)$ . From this, we have to make subject variable x and y. Then substituting these values of x and y in the given equation $x+y=4$ . Thus, we will get the answer.
Complete step-by-step answer:
equation $x+y=4$ . Thus, we will get the answer.
Here, we are given one end of diameter i.e. suppose let’s take $A\left( 3,3 \right)$ , other end of diameter as $B\left( x,y \right)$ and taking centre of circle as O, we get diagram as:
Here, O is midpoint of diameter AB so on finding midpoint of $B\left( x,y \right)$ $O\left( a,b \right)$ we get,
$a=\dfrac{3+x}{2}$
$2a=3+x$
Making x as subject variable, we get
$x=2a-3$ ………………………….(1)
Similarly, finding point b we get
$b=\dfrac{3+y}{2}$
$2b=3+y$
Making y as subject variable, we get
$y=2b-3$ …………………………….(2)
Now, substituting value of equation (1) and (2) into equation $x+y=4$ given in question, we get
$2a-3+2b-3=4$
$2a+2b-6=4$
On further solving, we get
$2a+2b=10$
Taking 2 Common from LHS and dividing it on RHS, we get
$a+b=5$ …………………………..(3)
So, again substituting values of a, b as x, y we get
$x+y=5$
Thus, the locus of the centre of the circle for which one end of the diameter is $\left( 3,3 \right)$ while the other end lies on the $x+y=4$ is $x+y=5$ .
Hence, option (b) is correct.
Note: Another method for solving this problem is taking centre point $O\left( a,b \right)$ , $A\left( 3,3 \right)$ and any point on the line $x+y=4$ by $\left( x,4-x \right)$ .
So, finding midpoints we get $a=\dfrac{3+x}{2}$ , $b=\dfrac{3+\left( 4-x \right)}{2}=\dfrac{7-x}{2}$
On further simplification and equating each other, we get
$2a-3-x=2b-7+x$
Thus, on further simplification we get $2a+2b=10$ and hence get the same answer as $x+y=5$ .
