Consider the equation of circle with centre at origin and radius . Then equation of such a circle is written as

Now equation of a line is represented by

To find the point where the line (ii) and touches the circle (i), using the method of solving simultaneous equation in which one equation is in quadratic and other is in linear form, so take value of from equation (ii) and putting this value in equation (i) i.e. equation of circle as follows

Since equation (iii) is a quadratic equation in , and compare this equation with the standard quadratic equation , to obtained the coefficients of and . Using these values in the discriminant i.e. .

If the discriminant is equal than zero i.e. , then equation (iii) will have equal roots and the line will intersect the circle at one point, i.e. it will be tangent to the circle as shown in the given diagram.

Thus the line is a tangent to the circle if

This is the condition for a line (ii) to be tangent to circle (i). Putting these values of in (ii), we have the tangents to the circle. Thus, the tangents to the circle are

__NOTE__**:**

**(1) **The condition **(iv) **is valid just for that circle whose centre is origin and for a circle with arbitrary centre this condition is not valid.

**(2) **The lines **(v)** are tangents to the circle whose centre is origin and for a circle with arbitrary centre these lines may not be the valid. ** **