# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2016 | June | Q#5

**Question**

A circle with center C(5,-3) passes through the point A(-2,1).

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

**b. **Given that AB is a diameter of the circle, find the coordinates of the point B.

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

**d. **The point T lies on the tangent to the circle at A such that AT=4. Find the length of CT.

**Solution**

**a.
**

We are given that circle has centre C(5,-3) and passes through the point A(-2,1).

Expression for a circle with center at and radius is;

We have coordinates of the center of the circle but we need to find radius.

It is evident that radius of the circle is AC with C(5,-3) and A(-2,1).

Expression to find distance between two given points and is:

Hence, radius of circle is;

Therefore, equation of the circle is;

**b.
**

We are given that circle has centre C(5,-3) and passes through the point A(-2,1).

It is also given that point B(x,y) is such that AB is diameter, therefore, point B also lies on the circle.

We are required to find the coordinates of point B.

We translate the given information into a diagram as shown below.

It is evident that both vertical and horizontal distance from A to C must be equal to both vertical and horizontal distance from C to B.

Therefore;

Since both vertical and horizontal distance from A to C must be equal to both vertical and horizontal distance from C to B, we can extend from C to B by extending both distances to corresponding coordinates of C.

For x-coordinate of point B, we need to move 7 units along positive x-axis away from point C, therefore, horizontal distance is added to x-coordinate of point C.

Similarly, for y-coordinate of point B, we need to move 4 units along negative y-axis away from point C, therefore, vertical distance is subtracted to y-coordinate of point C.

Hence, coordinates of point B(12,-7).

**c.
**

We are required to find equation of the tangent to the circle at point A(-2,1).

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(-2,1).

Therefore, we need slope of tangent.

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

Radius is always perpendicular to tangent.

Therefore;

Hence, first we need to find slope of radius of the circle AC.

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

We are given that C(5,-3) and A(-2,1).

Therefore;

Now, with coordinates of point on the tangent A(-2,1) and slope of tangent , we can write equation of the tangent.

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

**d. **

As we have seen in (c) that radius is always perpendicular to tangent, it is also evident that triangle ATC is a right-angled triangle. Therefore, we can apply Pythagorean Theorem.

Pythagorean Theorem

For triangle QCD;

We have found that AC=radius= and we are given that , hence;

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