A tangent drawn from point (4, 0) to circle x^2 + y^2 = 8 touches it at a point A in first quadrant.

English

Gujarati

Hindi

10-1.Circle and System of Circles

normal

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

$(2, - 2)$

B

$(- 2, 2)$

C

$\left( { - \,2\sqrt 2 \,\,,\,\,0} \right)$

D

$(A)$ or $(B)$ both

Solution

Equation of tangent through $(4, 0)$ is $x + y = 4$ . Point $'A'$ is $(2, 2)\,\, \Rightarrow \,\,A, B$

Standard 11

Mathematics

Share

Similar Questions

The point at which the normal to the circle ${x^2} + {y^2} + 4x + 6y – 39 = 0$ at the point $(2, 3)$ will meet the circle again, is

medium

The line $x\cos \alpha + y\sin \alpha = p$will be a tangent to the circle ${x^2} + {y^2} – 2ax\cos \alpha – 2ay\sin \alpha = 0$, if $p = $

medium

The equation of the tangent to the circle ${x^2} + {y^2} – 2x – 4y – 4 = 0$ which is perpendicular to $3x – 4y – 1 = 0$, is

medium

The equations of the tangents to the circle ${x^2} + {y^2} – 6x + 4y = 12$ which are parallel to the straight line $4x + 3y + 5 = 0$, are

medium

A circle with centre $'P'$ is tangent to negative $x$ & $y$ axis and externally tangent to a circle with centre $(-6,0)$ and radius $2$ . What is the sum of all possible radii of the circle with centre $P$ ?

normal