Let the abscissae of the two points P and Q on a circle be the roots of x2 - 4x - 6 = 0 and the ordinates of P and Q be the roots y2 + 2y - 7 = 0. If PQ¯ is a diameter of the circle x2 + y2 + 2ax + 2by + c = 0 then the value of (a + b - c) is

P=x1, y1, Q=x2, y2x2−4x−6=x−x1x−x2→1& y2+2y−7=y−y1y−y2→2Circle PQ¯ as diameter isx−x1x−x2+y−y1y−y2=0⇒ x2+y2−4x+2y−13=0→3Comparing, a=−2, b=1, c=−13→ a+b−c=−2+1+13=12

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Let the abscissae of the two points P and Q on a circle be the roots of x²- 4x - 6 = 0 and the ordinates of P and Q be the roots y²+ 2y - 7 = 0. If $\bar{P Q}$ is a diameter of the circle x²+ y²+ 2ax + 2by + c = 0 then the value of (a + b - c) is

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answer is 12.

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Detailed Solution

$\begin{array}{l} P = (x_{1}, y_{1}), Q = (x_{2}, y_{2}) \\ x^{2} - 4 x - 6 = (x - x_{1}) (x - x_{2}) \to (1) \\ & y^{2} + 2 y - 7 = (y - y_{1}) (y - y_{2}) \to (2) \end{array}$