Equation of the circle on the latusrectum of y 2-8 x as ends of diameter isA- x2-y2-4 y-12-0B- x2-y2-4 x-12-0C- x2-y2-4 y-16-0D- x2-y2-6 x-12-0

The correct option is A x2+y2 4y 12 =0 y^2=8x ex a=2 L = (2,4) L^1 = (2, 4)} Eqn of circle is (x 2)(x 2)+(y 4)(y+4)=0 ⇒ x^2+y^2 4x 12=0 ...

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Standard XII

Mathematics

Focii of Ellipse

Equation of t...

Question

Equation of the circle on the latusrectum of $y^{2} = 8 x$ as ends of diameter is

x²+y²-4y+16=0

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x²+y²-4y-12 =0

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x²+y²+4x+12=0

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x²+y²-6x-12 =0

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Solution

The correct option is B
x²+y²-4y-12 =0

$y^{2} = 8 x a = 2$
$\begin{matrix} L = (2, 4) L^{1} = (2, - 4) \end{matrix}}$
Eqn of circle is
(x-2)(x-2)+(y-4)(y+4)=0
$\Rightarrow x^{2} + y^{2} - 4 x - 12 = 0$

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Equation of the circle on the latusrectum of y2=8x as ends of diameter is

The correct option is B x2+y2-4y-12 =0 y2=8xa=2 L=(2,4)L1=(2,−4)} Eqn of circle is (x-2)(x-2)+(y-4)(y+4)=0 ⇒x2+y2−4x−12=0

Equation of the circle on the latusrectum of $y^{2} = 8 x$ as ends of diameter is

The correct option is B
x²+y²-4y-12 =0

$y^{2} = 8 x a = 2$
$\begin{matrix} L = (2, 4) L^{1} = (2, - 4) \end{matrix}}$
Eqn of circle is
(x-2)(x-2)+(y-4)(y+4)=0
$\Rightarrow x^{2} + y^{2} - 4 x - 12 = 0$