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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(x - 2)^{2} + y^{2} = 9$
Focus of parabola $y^{2} = 8 x$ is (2,0). Equation of circle with centre (2,0) is
$(x - 2)^{2} + y^{2} = r^{2}$
Let AB is common chord and Q is mid point i.e. (1,0)
$A Q^{2} = y^{2} = 8 x = 8 \times 1 = 8$
$∴ r^{2} = A Q^{2} + Q S^{2} = 8 + 1 = 9$
So required circle is $(x - 2)^{2} + y^{2} = 9$

Suggest Corrections

Similar questions

Q. From the focus of the parabola,

y^{2} = 8 x

as centre, a circle is described so that a common chord of the curve is equidistant from the vertex & focus of the parabola. The equation of the circle is

Q. From the focus of the parabola

y^{2} = 2 p x

as centre a circle is described so that a common chord of the curves is equidistant from the vertex and the focus of the parabola. Find the equation of the circle.

Q. Assertion :Length of the common chord of the parabola

y^{2} = 8 x

and the circle

x^{2} + y^{2} = 9

is less than the length of the latus rectum of the parabola. Reason: If the vertex of a parabola lies at the point

(a, 0)

and the directrix is

x + a = 0

, then the focus of the parabola is at the point

(2 a, 0)

Q. Let

S

be the focus of the focus of the parabola

y^{2} = 8 x

and

P Q

be the common chord of the circle

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

and the given parabola. The area of

△ P Q S

Equation of the parabola whose axis is y = x, distance from origin to vertex is $\sqrt{2}$ and distance from origin to focus is $2 \sqrt{2},$ is (Focus and vertex lie in Ist quadrant) :

Join BYJU'S Learning Program

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 -

The correct option is A (x−2)2+y2=9Focus of parabola y2=8x is (2,0). Equation of circle with centre (2,0) is(x−2)2+y2=r2Let AB is common chord and Q is mid point i.e. (1,0)AQ2=y2=8x=8×1=8∴r2=AQ2+QS2=8+1=9So required circle is (x−2)2+y2=9

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 -