From the focus of the parabola y 2=8 x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is -

Question

From the focus of the parabola y 2=8 x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is - (A) (x-2)2+y2=3 (B) (x-2)2+y2=9 (C) (x+2)2+y2=9 (D) x2+y2-4 x=0

Tardigrade Admin · Accepted Answer

Focus of parabola y2=8 x is (2,0). Equation of circle with centre (2,0) is (x-2)2+y2=r2 AB is common chord Q is mid point i.e. (1,0) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/75f0b924ab1420162f35f48e44253759-.png" /> AQ 2= y 2 where y 2=8 1=8 ∴ r 2= AQ 2+ QS 2=8+1=9 so circle is (x-2)2+y2=9

Q. From the focus of the parabola $y^{2} = 8 x$ as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is -

$(x - 2)^{2} + y^{2} = 3$

$(x - 2)^{2} + y^{2} = 9$

$(x + 2)^{2} + y^{2} = 9$

$x^{2} + y^{2} - 4 x = 0$

Focus of parabola $y^{2} = 8 x$ is $(2, 0)$ . Equation of circle with centre $(2, 0)$ is $(x - 2)^{2} + y^{2} = r^{2}$
$A B$ is common chord
$Q$ is mid point i.e. $(1, 0)$

$A Q^{2} = y^{2}$ where $y^{2} = 81 = 8$
$∴ r^{2} = A Q^{2} + Q S^{2} = 8 + 1 = 9$
so circle is $(x - 2)^{2} + y^{2} = 9$

Q. From the focus of the parabola y2=8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is -

(x−2)2+y2=3

(x−2)2+y2=9

(x+2)2+y2=9

x2+y2−4x=0