Question
Question: The locus of the centre of the circle for which one end of the diameter is \(\left( 3,3 \right)\) wh...
The locus of the centre of the circle for which one end of the diameter is (3,3) while the other end lies on the x+y=4 is
(a) x+y=3
(b) x+y=5
(c) x+y=7
(d) x+y=9
Solution
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 (a=2c+x,b=2d+y) . 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(3,3) , other end of diameter as B(x,y) and taking centre of circle as O, we get diagram as:
Here, O is midpoint of diameter AB so on finding midpoint of B(x,y) O(a,b) we get,
a=23+x
2a=3+x
Making x as subject variable, we get
x=2a−3 ………………………….(1)
Similarly, finding point b we get
b=23+y
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 (3,3) 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(a,b) , A(3,3) and any point on the line x+y=4 by (x,4−x) .
So, finding midpoints we get a=23+x , b=23+(4−x)=27−x
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 .