MHT CET · Maths · Circle
The co-ordinates of the mid-point of the chord cut off by theline \(2 x-5 y+18=0\)
by the circle \(x^{2}+y^{2}-6 x+2 y-54=0\) are
- A \((1,4)\)
- B \((2,4)\)
- C \((4,1)\)
- D \((1,1)\)
Answer & Solution
Correct Answer
(A) \((1,4)\)
Step-by-step Solution
Detailed explanation
Let \(A B\) be the chord
Let the mid point of chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\)
Here centre is \(\mathrm{O}=(3-1)\)
Here \(O M \perp A B\)
\(\therefore(\) slope of \(\mathrm{OM})(\) slope of \(\mathrm{AB})=-1\)
\(\left(\frac{\mathrm{k}+1}{\mathrm{~h}-3}\right)\left(\frac{2}{5}\right)=-1\)
\(\therefore 5 \mathrm{~h}+2 \mathrm{k}=13\)
Point \(M(h, k)\) lies on the \(2 x-5 y+18=0\)...(1)
\(\therefore 2 \mathrm{~h}-5 \mathrm{k}=-18\)...(2)
Solving equation (1) \& (2) we get \(\mathrm{h}=1, \mathrm{k}=4 \Rightarrow(1,4)\) is required point.
Let the mid point of chord be \(\mathrm{M}(\mathrm{h}, \mathrm{k})\)
Here centre is \(\mathrm{O}=(3-1)\)
Here \(O M \perp A B\)
\(\therefore(\) slope of \(\mathrm{OM})(\) slope of \(\mathrm{AB})=-1\)
\(\left(\frac{\mathrm{k}+1}{\mathrm{~h}-3}\right)\left(\frac{2}{5}\right)=-1\)
\(\therefore 5 \mathrm{~h}+2 \mathrm{k}=13\)
Point \(M(h, k)\) lies on the \(2 x-5 y+18=0\)...(1)
\(\therefore 2 \mathrm{~h}-5 \mathrm{k}=-18\)...(2)
Solving equation (1) \& (2) we get \(\mathrm{h}=1, \mathrm{k}=4 \Rightarrow(1,4)\) is required point.
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