# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2005 | June | Q#3

Question

A circle has center and radius 5. The point has coordinates .

a.   Write down the equation of the circle.

b.   Verify that point lies on the circle.

c.   Find the gradient of the line .

d.

i.  Find the gradient of the line which is perpendicular to .

ii. Hence find an equation for the tangent to the circle at point .

Solution

a.

Expression for a circle with center at and radius is; Therefore for a circle with center and radius 5;  b.

If a point lies on a circle then coordinates of point must satisfy the equation of the given circle.

From (a) we have equation of the circle; The point has coordinates ۔

Hence, substituting and coordinates of point in equation of the circle;    Therefore, point lies on the circle.

c.

Expression for slope (gradient) of a line joining points and ; For the given case we have and . Therefore;   d.

i.

If two lines are perpendicular (normal) to each other, then product of their slopes and is;  Therefore;  From (C) we have . Hence;   ii.

If we are required to find the equation of tangent to a circle at point :

ü Find the center of the circle

ü Find the gradient of ü Since radius and tangent are always perpendicular, find the gradient of tangent using Equation of the tangent can be written as .

We have coordinates of a point on the tangent to the circle and also from (d:i) we have slope of tangent (line perpendicular to radius CP). Therefore;        