Question

Find the equation of the normal to the circle x2+y2=40 at the point (6 2)

Answer 1

Since the tangent is perpendicular to the radius of the circle at the point (6,2), the normal, which is perpendicular to the tangent, must be parallel to the radius.

Center of the circle is (0,0)

So, slope of the normal is = (2-0)÷(6-0) = 2÷6 =1÷3

Equation of normal is: y - y1 = m(x-x1)

y - 2 = $\frac{1}{3}$(x-6)

y = $\frac{1}{3}$x