Question

The equations of the tangent at the point (0,0) to the circle ,making intercept of length 2a and 2b units on the coordinate axes,is/are

Answer 1

Let equation of circle be

      (x-a)² +(y-b)²=r².......(1)

It makes an intercept of length 2 a and 2 b units on the coordinate axes i.e it passes through (2 a, 0) and (0,2 b).

Equation of the circle becomes

a²+ b²=r², passing through (2 a,0).

and, a²+ b²=r²,passing through (0,2 b).

Equation 1 is

x² -2 a x + a²+y² - 2 b y + b²=r²

x² + y²-2 a x - 2 b y=0

as you can see (0,0) passes through above circle.

Center of the circle = (a,b)

Equation of line passing through (0,0) and (a,b) is

→a y = b x→b x - a y =0

Equation of line perpendicular to above line is

a x + b y + k=0

Since it passes through (0,0)

⇒a × 0 + b×0 + k=0

⇒ k =0

The above line becomes , a x + b y=0, which is the equation of tangent at the point (0,0) to the given circle.