**Question:**

Find the equation of the circle which passes through and and has, for its tangent at , the line .

**Solution:**

Recall that the equation of a circle is , where is the centre of the circle, and is the radius of the circle.

Substituting into the equation, we get:

— Eqn (1)

Substituting , we get:

— Eqn (2)

Equating Eqn (1) and Eqn (2), we get

which reduces to

— Eqn (3)

after simplification.

Now, we rewrite the equation of the tangent as (make y the subject)

Hence, the gradient of the normal is

Let the equation of the normal be

Substitute in we get

Hence

Thus equation of normal is

Since the normal will pass through the centre we have

— Eqn (4)

Finally, we equate Eqn (3) and Eqn (4),

Substituting back into Eqn (1), we get

Hence the equation of the circle is: