# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2012 | June | Q#6

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**Question**

The circle with centre C(5,8) touches the y-axis, as shown in the diagram.

**a. **Express the equation of the circle in the form

**b. **

** i. **Verify that the point A(2,12) lies on the circle.

** ii. **Find an equation of the tangent to the circle at the point A, giving your answer in the form , where s, t and u are integers.

**c. **The points P and Q lie on the circle, and the mid-point of PQ is M(7,12).

** i. **Show that the length of CM is , where n is an integer.

** ii. **Hence find the area of triangle PCQ.

**Solution**

**a.
**

We are given that circle has centre C(5,8) and touches the y-axis.

It is evident from the diagram and given fact that circle touches the y-axis, radius of the circle is equal to 5 i.e. equal to distance from y-axis to center of the circle.

Expression for a circle with center at and radius is;

Therefore;

**b.
**

i.

If a point lies on the circle then distance of point from the circle must be equal to radius of the circle.

We are given coordinates of a point A(2,12) and center of the circle C(5,8).

Let’s find distance between the two points.

Expression to find distance between two given points and is:

Therefore;

Since is equal to radius of circle as found in (a), point A lies on the circle.

ii.

We are required to find equation of tangent to the circle at point A(2,12).

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 A(2,12). Therefore, we need slope of tangent to write its equation.

Radius and tangent of the circle are always perpendicular.

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

For the given case;

Therefore, if we have slope of radius of the circle to which given tangent is perpendicular, we can find slope of the tangent.

We are looking for slope of tangent to the circle at point A therefore, we need slope of radius AC.

We are given coordinates of a point A(2,12) and center of the circle C(5,8).

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

Therefore;

Hence;

Now we can write equation of the tangent to circle at point A.

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

**c. **

i.

We are required to find the length of CM.

Expression to find distance between two given points and is:

We are given coordinates of a point M(7,12) and center of the circle C(5,8).

Let’s find distance between the two points.

ii.

Consider the diagram below.

We are required to find the area of triangle PCQ.

Expression for the area of the triangle is;

It is evident that PQ is the base of the triangle PCQ. Since both points P and Q lie on the circle PQ is a chord.

Perpendicular from center of the circle bisects the chord.

Since point M is the mid-point of PQ, CM is perpendicular to PQ and therefore, CM is height of the triangle PCQ.

Therefore;

We have already found in (c:i) that . Therefore, we need to find the length PQ.

It is evident from the diagram that;

Since M is the mid-point of PQ;

Therefore;

Consider the triangle CMP.

Pythagorean Theorem

It is evident from the diagram that;

We have already found in (c:i) that . Therefore;

Therefore;

Hence;

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