MHT CET · Maths · Circle
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
- A \(x+y=1\)
- B \(2(x-y)=5\)
- C \(2 x+2 y=5\)
- D None of these
Answer & Solution
Correct Answer
(C) \(2 x+2 y=5\)
Step-by-step Solution
Detailed explanation
Let the other end be \((t, 3-t)\).
So, the equation of the variable circle is
\((x-1)(x-t)+(y-1)(y-3+t)=0\)
\(\Rightarrow x^{2}+y^{2}-(1+t) x-(4-t) y+3=0\)
\(\therefore\) The centre \((\alpha, \beta)\) is given by
\(
\alpha=\frac{1+t}{2}, \beta=\frac{4-t}{2}
\)
\(
\Rightarrow 2 \alpha+2 \beta=5
\)
Hence, the locus is \(2 x+2 y=5\)
So, the equation of the variable circle is
\((x-1)(x-t)+(y-1)(y-3+t)=0\)
\(\Rightarrow x^{2}+y^{2}-(1+t) x-(4-t) y+3=0\)
\(\therefore\) The centre \((\alpha, \beta)\) is given by
\(
\alpha=\frac{1+t}{2}, \beta=\frac{4-t}{2}
\)
\(
\Rightarrow 2 \alpha+2 \beta=5
\)
Hence, the locus is \(2 x+2 y=5\)
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