The x-coordinates of two points $A$ and $B$ are roots of equation $x^{2} + 2 x - a^{2} = 0$ and y-coordinate are roots of equation $y^{2} + 4 y - b^{2} = 0$ . Then, equation of the circle which has diameter $A B$ , is

{(x - 1)}^{2} + {(y - 2)}^{2} = 5 + a^{2} + b^{2}

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{(x + 1)}^{2} + {(y - 2)}^{2} = \sqrt{5 + a^{2} + b^{2}}

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{(x + 1)}^{2} + {(y + 2)}^{2} = a^{2} + b^{2}

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{(x + 1)}^{2} + {(y + 2)}^{2} = 5 + a^{2} + b^{2}

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Solution

The correct options are
A ${(x - 1)}^{2} + {(y - 2)}^{2} = 5 + a^{2} + b^{2}$
C ${(x + 1)}^{2} + {(y + 2)}^{2} = 5 + a^{2} + b^{2}$
$x^{2} + 2 x - x^{2} = 0 \to x_{1}, x_{2}$
$= (x - x_{1}) (x - x_{2}) = 0$ (roots of eqn)
and
$y^{2} + 4 y - b^{2} = 0 \to y_{1}, y_{2}$
$= (y - y_{1}) (y - y_{2}) = 0$ (roots of eqn)
$A \equiv (x_{1}, y_{1})$ $B \equiv (x_{2}, y_{2})$
the eqn of circle with diameter $A B$ is given by $\Rightarrow$
$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$
$\Rightarrow x^{2} + 2 x - a^{2} + y^{2} + 4 y - b^{2} = 0$
$\Rightarrow (x^{2} + 2 x + 1) + (y^{2} + 4 y + 4)^{2} = a^{2} + b^{2} + 1 + 4$
$\Rightarrow (x + 1)^{2} + (y + 2)^{2} = a^{2} + b^{2} + 5$

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Similar questions

The equation of the circle with center (– a, – b) and radius √a² – b² is:

(x - 1)^{2} + y^{2} = a^{2}

(x + 2)^{2} + y^{2} = b^{2}

(x - a)^{2} + y^{2} = a^{2}, x^{2} + (y - b)^{2} = b^{2}

The equation of the common chord of the circles $(x - a)^{2} + (y - b)^{2} = c^{2}$ and $(x - b)^{2} + (y - a)^{2} = c^{2}$ is

(x - a)^{2} + (y - b)^{2} = c^{2}, (x - b)^{2} + (y - a)^{2} = c^{2}, (a \neq b)

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