The point $$R (4,10)$$ lies on the curve $$C: y = x^2 - 6x + 18.$$ The tangent and normal to $$C$$ at $$R$$ meets the Y-axis at points $$P$$ and $$Q$$ respectively. A circle passes through the points $$P,Q$$ and $$R$$. The radius of this circle is

$$\ \frac{\ dy}{dx}\ =\ 2x-6$$

Slope of the tangent at R(4, 10) = 2

A normal will be perpendicular to tangent, so the slope of the normal = $$\ \frac{\ -1}{2}$$

Equation of the tangent at R(4, 10) = 

y-10 = 2(x-4)

y=2x+2

The above line intersects Y-axis at P (0,2)

Equation of the normal at R(4, 10) =

y-10 = $$\ \frac{\ -1}{2}$$ (x-4)

2y=-x+24

The above line intersects Y-axis at Q (0,12)

Let the equation of the  circle be $$x^2+y^2+2gx+2fy+c=0$$ with centre (-g, -f)

Now we have to find the equation of the circle passing through P (0, 2), Q(0, 12), R(4, 10)

On substituting these values in the equation of the circle, we get g = 0, f = -7, c = 24

Radii of the circle = $$\sqrt{\ \left(g^2+f^2-c\right)}$$

= 5