AP EAMCET · Maths · Circle
If \(x-y+1=0\) meets the circle \(x^2+y^2+y-1=0\) at \(A\) and \(B\), then the equation of the circle with \(A B\) as diameter is
- A \(2\left(x^2+y^2\right)+3 x-y+1=0\)
- B \(2\left(x^2+y^2\right)+3 x-y+2=0\)
- C \(2\left(x^2+y^2\right)+3 x-y+3=0\)
- D \(x^2+y^2+3 x-y+1=0\)
Answer & Solution
Correct Answer
(A) \(2\left(x^2+y^2\right)+3 x-y+1=0\)
Step-by-step Solution
Detailed explanation
Given that \(\Rightarrow x^2+(x+1)^2+x+1-1=0 \quad\) [from Eq. (i)] \(\begin{array}{lc} \Rightarrow & 2 x^2+3 x+1=0 \\ \Rightarrow & (2 x+1)(x+1)=0 \\ \Rightarrow & x=-\frac{1}{2},-1 \text { and } y=\frac{1}{2}, 0 \end{array}\) \(\therefore\) Point of…
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