Question

The points A and B have coordinates (-8,-8) and (12,2) respectively AB is the diameter of circle C
(a) find an equation for C
The point ( 4,8) also lies on C
find an equation of the tangent to c at the point (4,8) giving your answer in for of ax + by + c= 0

Answer 1

(a) The equation of a regular circle with radius R is

$$R^2=(x-a)^2+(y-b)^2.$$

Substitute the point we are given:

$$R^2=(-8-a)^2+(-8-b)^2,\\ R^2=(12-a)^2+(2-b)^2.$$

We have three undefined values. However, we know that the distance between these two given points is the diameter, so, the radius is

$$R=\frac D2=\frac{\sqrt{(-8-12)^2+(-8-2)^2}}{2}=5\sqrt5.$$

Substitute and find a and b:

$$125=(-8-a)^2+(-8-b)^2,\\ 125=(12-a)^2+(2-b)^2.\\ a=2\\ b=-3$$

So, the equation of the circle is

$$125=(x-2)^2+(y+3)^2.$$

(b) To find the equation of a tangent, express y in terms of x and find the derivative:

$$y'=(\sqrt{-x^2+4x+121}-3)'=\\\space\\ =\frac{2-x}{\sqrt{-x^2+4x+121}}.$$

Substitute x-value to find the gradient:

$$k=y'(4)=-\frac{2}{11}.$$

The equation is

$$y-y_1=k(x-x_1),\\ y-8=-\frac{2}{11}(x-4),\\\space\\ \frac{2}{11}x+y-\frac{96}{11}=0.$$