How to find the equation of the circle using the tangent or derivative.

Question

Find the equation of the tangent line to the curve y = x² at x = 1.

so I found this to be y=2x-1

 

the part I need help with is the circle equation:

 
b) This line is also tangent to a circle with center (28, 0) at x = 6. Find the equation of this circle.

 

I tried doing (x-28)2+y²=495 but that as wrong, so now I am a little confused on what exactly I am trying to do and find.

 

 

Answer 1

The tangent line is y = 2x-1.  When x = 6, y = 11.  So, the point (6,11) lies on the tangent line and also lies on the circle. 

 

Equation of circle:  (x-28)²+(y-0)² = r²

 

Since (6,11) is on the circle, we have (6-28)²+(11-0)² = r^2.

 

Therefore, r² = 605

 

Equation of circle:  (x-28)²+y² = 605

Answer 2

Make a sketch. put C(28,0) as center; draw radius from C that is perpendicular to the tangent; this radius will have slope -½ because it's perpendicular to tangent line that has slope of 2 (remember 'negative reciprocal'?).

 

So you can write the equation of that radius, and you can find where the radius intersects the tangent by solving the system of equations for these two lines.

 

When you get that intersection point, you can calculate the distance from it to C, and thus obtain r for the circle.

The rest is easy.