Question

Tangents PA and PB are drawn to the circle x²+y²=4, then the locus of the point P if the triangle PAB is equilateral, is equal to -

A. x²+y²=16
B. x²+y²=8
C. x²+y²=64
D. x²+y²=32

Answer 1

The locus of the point P, if the triangle PAB is equilateral, is x²+y²=8. To find this, we can start by finding the coordinates of point A and point B using the circle equation, and then use the fact that the triangle is equilateral to find the coordinates of point P.

Step-by-step explanation:

The locus of the point P, if the triangle PAB is equilateral, is given by the equation x²+y²=8 (option B).

To solve this problem, we can start by finding the coordinates of point A and point B using the circle equation. Substituting x=2 and y=0 into the equation x²+y²=4, we find that point A is (2,0).

Similarly, substituting x=-2 and y=0, point B is (-2,0).

Now, we can use the fact that the triangle PAB is equilateral to find the coordinates of point P.

Since PA and PB are tangents to the circle, they are perpendicular to the radius of the circle at points A and B.

Therefore, the distance from the center of the circle to point P is also 4.

Since the radius of the circle is 2, the coordinates of point P are (0,4).

Plugging these coordinates into the equation of a circle, we find that 0²+4²=8.

Therefore, the locus of point P is x²+y²=8.