# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2013 | January | Q#7

Question

A circle with centre C(-3,2) has equation

a.   Find the y-coordinates of the points where the circle crosses the y-axis.

b.   Find the radius of the circle.

c.   The point P(2,5) lies outside the circle.

i.       Find the length of CP, giving your answer in the form  , where n is an integer.

ii.       The point Q lies on the circle so that PQ is a tangent to the circle. Find the length of PQ.

Solution

a.

We are given that circle has centre C(5,8) and has equation;

We are required to find the y-intercept of the circle.

The point  at which curve (or line) intercepts y-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

Therefore;

Now we have two options.

b.

Expression for a circle with center at  and radius  is;

We are given that circle has equation;

We can write the equation of the circle in standard form.

We can rearrange the given equation and compare the given terms with standard expression.

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

 Therefore we can deduce that;

Hence we can write;

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

Expression for a circle with center at  and radius  is;

Comparing this equation with standard form of the equation of circle, we can write;

Hence, radius of the circle is 5.

c.

i.

We are required to find the distance CP.

Expression to find distance between two given points  and is:

We have coordinates of both points P(2,5) and C(-3,2). Therefore;

ii.

Radius is always perpendicular to tangent.

Therefore, PQ and radius of circle r are perpendicular to each other.

Hence, PQC make a right angle triangle.

Pythagorean Theorem

We have found above that  and .