Question

The locus of the centre of the circle for which one end of a diameter is $\left(1,1\right)$ while the other end is in the line $x+y=3$ 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 answer:

Let the center of the circle be $C(h,k)$ and the other end of the diameter be $A(a,b)$

Since $C(h,k)$ is the midpoint of $A(a,b)$ and $\left(1,1\right)$, by midpoint formula,

so, $h=\frac{a+1}{2},k=\frac{b+1}{2}$

$$\begin{gathered} 2h=a+1,2k=b+1 \\ a=2h-1,b=2k-1 \end{gathered}$$

but $A(a,b)$ lies on $x+y=3$

$$\begin{gathered} \Rightarrow a+b=3 \\ \Rightarrow 2h-1+2k-1=3 \\ \Rightarrow 2h+2k-2=3 \\ \Rightarrow 2h+2k=5 \end{gathered}$$

Substitute $h=x,y=k$

$\Rightarrow 2x+2y=5$

Hence, option C is correct.