Question

The locus of the point of intersection of the tangents at the extremities of a chord of the circle x²+y²=a² which touches the circle x²+y²−2ax=0 passes through the point

a. (a/2,0)
b. (0,a/2)
c. (o,a)
d. (a,0)

Answer 1

The locus of the point of intersection of the tangents at the extremities of a chord of a circle can be found by considering the properties of the circle and the given equation. The equation of the line passing through the point of contact and perpendicular to each of the tangent lines will pass through the point (a/2, 0) as given in option a.

Step-by-step explanation:

The locus of the point of intersection of the tangents at the extremities of a chord of a circle can be found by considering the properties of the circle and the given equation.

First, we know that a chord of a circle that touches the circle is perpendicular to the radius at the point of contact. In this case, the given equation represents the circle x²+y²−2ax=0, which is a circle with center at the origin (0,0) and radius a.

The point of contact of the chord with the circle can be found by substituting the equation of the chord, which is y = mx, into the equation of the circle. By solving for x and y, we can find the coordinates of the point of contact. From there, we can find the equation of the tangent lines at the extremities of the chord by using the slope-intercept form of a line, y = mx + b.

Finally, we can find the locus of the point of intersection of the tangents by finding the equation of the line passing through the point of contact and perpendicular to each of the tangent lines. This line will pass through the point (a/2, 0) as given in option a.