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If one end of the diameter is $(1,1)$ and the other end lies on the line $x+y=3$, then locus of centre of circle is
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2994 Upvotes
Verified Answer
The correct answer is:
$2 x+2 y=5$
Let the other end be $(t, 3-t)$.
So, the equation of the variable circle is
$$
\begin{array}{l}
(x-1)(x-t)+(y-1)(y-3+t)=0 \\
\Rightarrow x^{2}+y^{2}-(1+t) x-(4-t) y+3=0
\end{array}
$$
$\therefore$ The centre $(\alpha, \beta)$ is given by
$$
\alpha=\frac{1+t}{2}, \beta=\frac{4-t}{2}
$$
$$
\Rightarrow \quad 2 \alpha+2 \beta=5
$$
Hence, the locus is $2 x+2 y=5$
So, the equation of the variable circle is
$$
\begin{array}{l}
(x-1)(x-t)+(y-1)(y-3+t)=0 \\
\Rightarrow x^{2}+y^{2}-(1+t) x-(4-t) y+3=0
\end{array}
$$
$\therefore$ The centre $(\alpha, \beta)$ is given by
$$
\alpha=\frac{1+t}{2}, \beta=\frac{4-t}{2}
$$
$$
\Rightarrow \quad 2 \alpha+2 \beta=5
$$
Hence, the locus is $2 x+2 y=5$
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