The abscissae of two points A and B are roots of x2 + 4x – 45 = 0 and ordinates of A, B are roots of y2 – 7y + 12 = 0. Then centre of circle for which A, B as diameter is

x1+x2=−4;x1x2=−45;y1+y2=7;y1y2=12eqx−x1x−x2+y−y1y−y2=0 ⇒Centre=x1+x22,y1+y22=-42,72

The abscissae of two points A and B are roots of x2 + 4x – 45 = 0 and ordinates of A, B are roots of y2 – 7y + 12 = 0. Then centre of circle for which A, B as diameter is

x1+x2=−4;x1x2=−45;y1+y2=7;y1y2=12eqx−x1x−x2+y−y1y−y2=0 ⇒Centre=x1+x22,y1+y22=-42,72

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The abscissae of two points A and B are roots of x² + 4x – 45 = 0 and ordinates of A, B are roots of y² – 7y + 12 = 0. Then centre of circle for which A, B as diameter is

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(2, 7/2)

(–2, 7/2)

(2, –7/2)

(–2, –7/2)

answer is B.

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Detailed Solution

$\begin{array}{l} x_{1} + x_{2} = - 4; x_{1} x_{2} = - 45; y_{1} + y_{2} = 7; y_{1} y_{2} = 12 \\ eq (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0 \end{array} \Rightarrow C e n t r e = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (\frac{- 4}{2}, \frac{7}{2})$