From the focus of the parabola $y^{2} = B x$ , tangents are drawn to the circle $(x - 6)^{2} + y^{2} = 4$ . Then, the equation of circle through the locus and points of contact of the tangents is

x^{2} + y^{2} - 8 x + 12 = 0

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x^{2} + y^{2} + 6 x - 12 = 0

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x^{2} + y^{2} + 8 x - 12 = 0

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

x^{2} + y^{2} - 6 x - 12 = 0

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

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Solution

The correct option is A $x^{2} + y^{2} - 8 x + 12 = 0$
AB is chord of contact of tangents drawn from focus P(2,0) to the circle
$x^{2} + y^{2} - 12 x + 32 = 0$

Its equation is
$x^{2} + y .0 - 6 (x + 2) + 32 = 0$
or x−5=0,x=5
Then by S+λP=0
the equation of circle is
$x^{2} + y^{2} - 12 x + 32) + λ (x - 5) = 0$

It passes through (2,0)
∴12−3λ=0 or λ=4.
$x^{2} + y^{2} - 8 x + 12 = 0$

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Q. Tangents are drawn to the circle

x^{2} + y^{2} + 6 x + 4 y - 12 = 0

from origin. Determine the equation of the circle passing through the points of contact of the tangents and the origin.

Q. Find the equation of the circle passing through the points of intersection of the circles

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

and

x^{2} + y^{2} + 6 x + 4 y - 12 = 0

and intersection the circle

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

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The equation of the common tangent to the circles $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ at their point of contact.

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From the focus of the parabola y2=Bx, tangents are drawn to the circle (x−6)2+y2=4. Then, the equation of circle through the locus and points of contact of the tangents is

From the focus of the parabola $y^{2} = B x$ , tangents are drawn to the circle $(x - 6)^{2} + y^{2} = 4$ . Then, the equation of circle through the locus and points of contact of the tangents is