AP EAMCET · Maths · Circle
If the mid point of the chord intercepted by the circle \(x^2+y^2-8 x+10 y+5=0\) on the line \(2 x+y+2=0\) is \((\mathrm{h}, \mathrm{k})\), then \(\mathrm{k}+4 \mathrm{~h}=\)
- A \(2\)
- B \(0\)
- C \(1\)
- D \(-1\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
Given the equation of circle \(\begin{aligned} & x^2+y^2-8 x+10 y+5=0 \\ & \text { Centre }=(4,-5) \end{aligned}\) Now, equation of chord is \(2 \mathrm{x}+\mathrm{y}+2=0\) since \((\mathrm{h}, \mathrm{k})\) is mid point of chord \(\Rightarrow 2 \mathrm{~h}+\mathrm{k}+2=0\) ...…
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