# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | June | Q#5

Question

A circle with centre C has equation  .

a.   Write down:

i.       the coordinates of C;

ii.       the radius of the circle.

b.

i.       Verify that the point N(0,-2) lies on the circle.

ii.       Sketch the circle.

iii.       Find an equation of the normal to the circle at the point N.

c.   The point P has coordinates (2, 6).

ii.       Find the length of a tangent drawn from P to the circle.

Solution

a.

We are given equation of the circle as;

Expression for a circle with center at  and radius  is;

We can write the given equation of the circle in standard form as;

Comparing the equation of the circle with expression for circle;

i.

Coordinates of center of circle .

ii.

of circle is 5
.

b.

i.

If point N(0,-2) lies on the circle then distance of point P from center of the circle must be equal to  radius of the circle.

Expression to find distance between two given points  and is:

We have two points with coordinates  and . Therefore;

Hence point N(0,-2) lies on the circle.

ii.

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 5.

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

iii.

We are required to find an equation of the normal to the circle at the point N(0,-2).

The normal at a given point P on a circle with center  is the same as the radius .

Therefore we can write equation of the normal to the circle at point N(0,-2) from equation of the radius of the circle CN.

We have coordinates of two points on the radius CN as N(0,-2) and C(-3,2).

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

Therefore;

c.

i.

Expression to find distance between two given points  and is:

We have two points with coordinates  and . Therefore;

ii.

Radius is always perpendicular to tangent.

If we have point P(2, 6) outside circle and then a tangent is drawn from point P to the circle, say at point S(x,y), then PS will be always perpendicular to the radius of the circle CS.

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

Pythagorean Theorem

We know that  (from (a:ii)), and  (from C:i). Therefore;