Question

If one end of the diameter is $\left(1,1\right)$ and the other end lies on the line $x+y=3$, then the locus of the center of a circle is

• A. $x+y=1$
• B. $2\left(x–y\right)=5$
• C. $2x+2y=5$
• D. None of these

Answer 1

The correct option is C

$2x+2y=5$

Explanation for the correct option.

Step 1: Given information

Here one end of the diameter is given and it is given that other end is on the line with equation $x+y=3$.

Now, let the $x-\text{coordinate}$ of the other end be $t$ then its $y-\text{coordinate}$ will be $3-t$. So, the coordinates of the other end will be $\left(t,3-t\right)$.

As the coordinates of both the ends of the diameter are known then the equation of the variable circle can be written as,

$$\begin{gathered} \therefore \left(x-1\right)\left(x-t\right)+\left(y-1\right)\left(y-\left(3-t\right)\right)=0 \\ \Rightarrow x^2-xt-x+t+y^2-3y+yt-y+3-t=0 \\ \Rightarrow x^2+y^2-\left(1+t\right)x-\left(4-t\right)y+3=0 \end{gathered}$$

We know that the general form of equation of circle is $x^2+y^2+2gx+2fy+c=0$, where $\left(-g,-f\right)$ are the coordinates of the center of a circle.

After the comparison of general form of equation of circle and equation of variable circle we have,

$$\begin{gathered} 2g=1+t\dots \left(1\right) \\ 2f=4-t\dots \left(2\right) \end{gathered}$$

Step 2 : Add the equations $\left(1\right)\mathbf{}\mathbf{\text{and}}\mathbf{}\left(2\right)$

$2g+2f=5$

Therefore, the locus of the center of a circle is $2x+2y=5$.

Hence, the correct option is (C).