MHT CET · Maths · Circle
Equation of the chord of the circle \(x^2+y^2-4 x-10 y+25=0\) having mid-point \((1,2)\) is
- A \(-x+3 y=5\)
- B \(x+3 y=7\)
- C \(5 x+y=7\)
- D \(3 x+y=5\)
Answer & Solution
Correct Answer
(B) \(x+3 y=7\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& x^2+y^2-4 x-10 y+25=0 \Rightarrow \text { centre }=(2,5) \text { and } \\
& \text { radius }=\sqrt{4+25-25}=2
\end{aligned}
\)
Refer figure
Let \(\mathrm{M}(1,2)\) be the midpoint of chord
Slope of \(\mathrm{CM}=\frac{2-5}{1-2}=3\)
\(\therefore\) Slope of \(\mathrm{AB}=\frac{-1}{3}\)
Equation of \(\mathrm{AB}\) is \((\mathrm{y}-2)=\frac{-1}{3}(\mathrm{x}-1)\) i.e. \(\mathrm{x}+3 \mathrm{y}=7\)
\begin{aligned}
& x^2+y^2-4 x-10 y+25=0 \Rightarrow \text { centre }=(2,5) \text { and } \\
& \text { radius }=\sqrt{4+25-25}=2
\end{aligned}
\)
Refer figure
Let \(\mathrm{M}(1,2)\) be the midpoint of chord
Slope of \(\mathrm{CM}=\frac{2-5}{1-2}=3\)
\(\therefore\) Slope of \(\mathrm{AB}=\frac{-1}{3}\)
Equation of \(\mathrm{AB}\) is \((\mathrm{y}-2)=\frac{-1}{3}(\mathrm{x}-1)\) i.e. \(\mathrm{x}+3 \mathrm{y}=7\)
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