AP EAMCET · Maths · Circle
A circle \(S\) touches \(Y\)-axis at \((0,3)\) and makes an intercept of length 8 units on \(\mathrm{X}\)-axis. If the centre \(\mathrm{C}\) of the circle \(\mathrm{S}\) lies in the second quadrant, then the distance of \(\mathrm{C}\) from the point \((-2,-1)\) is
- A \(13\)
- B \(10\)
- C \(5\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
Equation of circle touches y-axis at \((0,3)\) is \((x-a)^2+(y-3)^2=a^2 \Rightarrow x^2+y^2-2 a x-6 y+9=a^2\) Since the length of intercept on \(\mathrm{x}\)-axis at this circle is 8 unit \(\therefore 2 \sqrt{a^2-9}=8 \Rightarrow a^2=25 \Rightarrow a= \pm 5\) Center lies in…
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