Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2009 | January | Q#7

Question

A circle with center C has equation a.   Express this equation in the form b.   Write down:

i.       the coordinates of C;

ii.       the radius of the circle

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

i.       Verify that point D lies on the circle.

ii.       Find an equation of the normal to the circle at the point D, giving your answer in the form , where m, n and p are integers.

d.

i.       A line has equation . Show that the x-coordinates of any points of intersection of  the line and the circle satisfy the equation ii.       Find the values of k for which the equation has equal roots. (5 marks)

iii.       Describe the geometrical relationship between the line and the circle when k takes either  of the values found in part (d)(ii).

Solution

a.

We have the algebraic formula;  We have the algebraic formula;  For the given case we can rearrange the given equation and compare the given terms with the  formula.   For terms containing For terms containing     Therefore, we can deduce that;   To complete the square we can add and subtract the deduced value of ;         b.

Expression for a circle with center at and radius is; Comparing the equation from (a) with expression for circle;

i.

Coordinatesof center of circle .

ii.

c.

i.

We are required to prove that the point D(7,-2) lies on the circle.

It is evident that if point D(7,-2) 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 (b:1) that center of the circle C(3,-5) and radius of the circle is .

Therefore, if point D(7,-2) lies on the circle;        Hence point D(7,-2) lies on the circle.

ii.

We are required to find an equation of the normal to the circle at the point D(7,-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 D(7,-2) from equation of the  radius of the circle CD.

We have coordinates of two points on the radius CD as C(3,-5) and D(7,-2).

Two-Point form of the equation of the line is; Therefore;            d.

i.

Since line and the circle intersect, we find the coordinates of points of intersection.

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have same values on both lines (or on the line and the curve).  Therefore, we can equate coordinates of both lines i.e. equate equations of both the lines (or the  line and the curve).

Equation of the line is; Equation of the circle from (a) is; Substituting value of from equation of line into equation of circle;     ii.

It is evident that given equation is a parabola; For a quadratic equation , the expression for solution is; Where is called discriminant.

If , the equation will have two roots.

If , the equation will have two identical/repeated roots.

If , the equation will have no roots.

According to the given condition, the given equation has equal roots. Therefore, for the given  equation;          Now we have two options.      Hence, given equation will have equal roots for and .

iii.

Hence, equation which satisfies x-coordinates of any points of intersection of the line and the circle,  will have equal roots for and .

Two equal or repeated roots in the solution of the equation which satisfies x-coordinates of any points of intersection of the line and the circle, depicts that there is only one point of intersection  of the line and the circle.

This indicates that line is tangent to the circle.