# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | June | Q#5

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Question

A circle with center C has equation

a.   Write down:

i.       the coordinates of C;

ii.       the radius of the circle

b.   The point D has coordinates (7,-2).

i.       Verify that the circle passes through the origin O.

ii.       Given that the circle also passes through the points (10,0) and (0,p), sketch the circle  and find the value of p.

c.   The point A(-7,-7) lies on the circle.

i.       Find the gradient of AC.

ii.       Hence find an equation of the tangent to the circle at point A, giving your answer in the  form , where a, b and c are integers.

Solution

a.

i.

Expression for a circle with center at  and radius  is;

Comparing this equation with given equation for circle;

i.

Coordinates of center of circle .

ii.

Radius of circle 13.

b.

i.

If circle passes through the origin then point O(0,0) lies on the circle . Therefore, we are required to  prove that the point O(0,0) lies on the circle.

It is evident that if point O(0,0) lies on the circle, then distance from center of the circle to this point  must be equal to radius of the circle.

Expression to find distance between two given points  and is:

We have found in (a:ii) that center of the circle C(5,-12) and radius of the circle is .

Therefore, if point O(0,0) lies on the circle;

Hence point O(0,0) lies on the circle.

ii.

We are given that the circle also passes through the points (10,0) and (0,p).

Let P(10,0) and Q(0,p), then CP and CQ area also both radii of the given circle and hence have  lengths equal to 13.

Let’s find coordinates of Q(0,p).

Expression to find distance between two given points  and is:

We have found in (a:ii) that center of the circle C(5,-12) and radius of the circle is .

Therefore, if point Q(0,p) lies on the circle;

Now we have two options.

It is evident that  is not possible because that will make coordinates of Q(0,0) which is origin  and we already have origin on the circle.

Hence;

We are required to sketch the circle.

ü Find the radius and coordinates of the center of the circle

From (a) we have found that coordinates of center of the circle are  and radius of circle is 13.

ü Indicate the center

ü Mark the four points which show the ends of the horizontal and vertical diameters

ü Draw the circle to pass through these four points

ü If any intercepts with the coordinate axes are integers, normally they should also be indicated

c.

i.

We are given that point A(-7,-7) lies on the circle and we are required to find the gradient of AC where C is the centre of the circle.

From (a:i) we have C(5,-12).

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

ii.

We are required to find equation of the tangent to the circle at point A.

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 are given coordinates of a point on the tangent A(-7,-7).

Therefore, we need slope of tangent.

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

We already know the slope of line CA  from (c:i) which is perpendicular to the tangent of  the circle at point A.

Therefore;

Now, with coordinates of point on the tangent A(-7,-7) and slope of tangent , we can write  equation of the tangent.

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

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