The area of the triangle formed by the tangents from the points $(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ and the line joining their points of contact is
The area of the triangle formed by the tangents from the points $(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ and the line joining their points of contact is
- A
$a{\rm{ }}\frac{{{{({h^2} + {k^2} - {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$
- B
$a{\rm{ }}\frac{{{{({h^2} + {k^2} - {a^2})}^{1/2}}}}{{{h^2} + {k^2}}}$
- C
$\frac{{{{({h^2} + {k^2} - {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$
- D
$\frac{{{{({h^2} + {k^2} - {a^2})}^{1/2}}}}{{{h^2} + {k^2}}}$
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A tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The co-ordinates of another point $B$ on the circle such that $l\, (AB) = 4$ are :
A tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The co-ordinates of another point $B$ on the circle such that $l\, (AB) = 4$ are :
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A tangent $P T$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ A common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ A possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$
Give the answer question $1$ and $2.$
A tangent $P T$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ A common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ A possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$
Give the answer question $1$ and $2.$
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Tangents are drawn from $(4, 4) $ to the circle $x^2 + y^2 - 2x - 2y - 7 = 0$ to meet the circle at $A$ and $B$. The length of the chord $AB $ is
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