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;

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