Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2014 | June | Q#7

 

Question

A circle with center C has the equation . The point A(3,-2) lies on the  circle.

a.   Express the equation of the circle in the form

b.    

                           i.       Write down the coordinates of C.

                         ii.       Show that the circle has radius , where n is an integer.

c.    

                    i.       Find the equation of the tangent to the circle at the point A, giving your answer in the form  , where p and q are integers.

                  ii.       The point B lies on the tangent to the circle at A and the length of BC is 6. Find the length  of AB.

Solution

a.
 

We are given equation of the circle with center C as;

Expression for a circle with center at  and radius  is;

We can write the given equation of the circle in standard for as follows.

We have the algebraic formula;

For the given case we can rearrange the given equation and compare the given terms with the  formula.

For terms containing  

For terms containing  

Therefore, we can deduce that;

To complete the square we can add and subtract the deduced value of ;

b.
 


i.
 

Expression for a circle with center at  and radius  is;

We can write the equation of circle obtained in (a) in standard form as;

Comparing the equation from (a) with expression for circle we get coordinates of center of circle .

 

                          ii.       

Radius of circle .

c.
 

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We already have coordinates of a point on the tangent to the circle at point A(3,-2). 

Therefore, we need slope of the tangent to the curve at point A(3,-2). 

Radius is always perpendicular to tangent.

If two lines are perpendicular (normal) to each other, then product of their slopes  and  is; 

Therefore, if we have slope of the radius of circle from center to point A, we can find slope of  tangent to the circle at point A. 

We now find slope of the radius of circle from center to point A.

Expression for slope (gradient) of a line joining points  and ;

We have coordinates of both point A(3,-2) and C(5,-6).

We know that;

Now with coordinates of point A(3,-2) and slope of tangent to the circle at this point  we can  write equation of this tangent to the circle.

Point-Slope form of the equation of the line is;

d.
 

Radius is always perpendicular to tangent.

If we have point B on tangent to the circle at point A that means point B is essentially outside the  circle because only point on both the circle and tangent is A whereas all other points on the tangent  are outside the circle.

If we have point B outside circle and then a tangent is drawn from point A to the circle, , then AB will  be always perpendicular to the radius of the circle CS.

This makes a right-triangle BAC with angle BAC=90o. We can apply Pythagoras theorem to the  triangle.

Pythagorean Theorem

We know that  (from (b:ii), and  (given).
Therefore;

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