AP EAMCET · Maths · Circle
A circle passing through origin cuts the coordinate axes is \(A\) and \(B\). If the straight line \(A B\) passes through a fixed point \(\left(x_1, y_1\right)\), then the locus of the centre of the circle is
- A \(\frac{x_1}{x}+\frac{y_1}{y}=1\)
- B \(x_1 y=x y_1\)
- C \(x y_1+y x_1=2\)
- D \(\frac{x_1}{x}+\frac{y_1}{y}=2\)
Answer & Solution
Correct Answer
(D) \(\frac{x_1}{x}+\frac{y_1}{y}=2\)
Step-by-step Solution
Detailed explanation
Circle through origin: \(x^2+y^2+2gx+2fy=0\). Intercepts on axes: \(A(-2g, 0)\), \(B(0, -2f)\). Equation of line AB: \(\frac{x}{-2g} + \frac{y}{-2f} = 1\) Line AB passes through \((x_1, y_1)\): \(\frac{x_1}{-2g} + \frac{y_1}{-2f} = 1\) Locus of center \((x,y)=(-g,-f)\):…
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