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AP EAMCET · Maths · Circle

If one end of diameter of the circle \(x^2+y^2-4 x-6 y+11=0\) is \((3,4)\), then the other end of the diameter is

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

Answer & Solution

Correct Answer

(C) \((1,2)\)

Step-by-step Solution

Detailed explanation

Given, circle \(x^2+y^2-4 x-6 y+11=0\) \(\Rightarrow \quad\) Centre \(=(2,3)\) One end diameter \(=(3,4)\) Let other end be \((h, k)\) So, \(\frac{h+3}{2}=2, \frac{k+4}{2}=3\) \(\Rightarrow h=1, k=2\)
From AP EAMCET
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