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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 \(\mathrm{k}+4 \mathrm{~h}=\)

  1. A \(2\)
  2. B \(0\)
  3. C \(1\)
  4. D \(-1\)
Verified Solution

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 \(\Rightarrow 2 \mathrm{~h}+\mathrm{k}+2=0\) ...…