Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2010 | June | Q#5

 

Question

A circle with centre  touches the y-axis, as shown in the diagram.

a.   Find the equation of the circle in the form

b.    

                     i.       Verify that the point  lies on the circle.

                  ii.       Find an equation of the normal to the circle at the point P.

                iii.       The mid-point of PC is M. Determine whether the point P is closer to the point M or to the  origin O.

Solution

a.
 

Expression for a circle with center at  and radius  is;

We are given that center of the circle has coordinates .

We need radius of the given circle to write its equation.

It is evident from the diagram that circle touches the y-axis. Therefore, horizontal distance from center of the circle is 5.

Therefore;

b.
 


i.
 

It is evident that if point  lies on the circle then distance between center of the circle and this  point must be equal to radius of the circle.

Expression to find distance between two given points  and is:

We have coordinates of both point  and . Therefore;

Hence, point P lies on the circle.


ii.
 

We are required to find the equation of a line which is normal to the circle at point .

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 normal to the curve and that is .

We need slope of the normal to the curve at point  to write its equation.

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

Since normal to the circle at point P is perpendicular to line (radius) CP, if we have slope of line CP  then we can find slope of the normal to the curve.

Let’s find slope of the line CP.

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

We have coordinates of both points  and . Therefore;

Therefore;

Now we can write equation of the normal.

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


iii.
 

We are given that mid-point of PC is M.

We need to find and compare distances MP and OP.

We know that CP is the radius of the circle as found in (b:i) and M is the mid-point of CP therefore;

Now we need to find OP.

Expression to find distance between two given points  and is:

We have coordinates of point  and

Therefore;

Since , therefore;

Hence point P is closer to point M than point O.

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