Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | June | Q#5
Question
A circle with centre C has equation .
a. Write down:
i. the coordinates of C;
ii. the radius of the circle.
b.
i. Verify that the point N(0,-2) lies on the circle.
ii. Sketch the circle.
iii. Find an equation of the normal to the circle at the point N.
c. The point P has coordinates (2, 6).
i. Find the distance PC, leaving your answer in surd form.
ii. Find the length of a tangent drawn from P to the circle.
Solution
a.
We are given equation of the circle as;
Expression for a circle with center at and radius
is;
We can write the given equation of the circle in standard form as;
Comparing the equation of the circle with expression for circle;
i.
Coordinates of center of circle .
ii.
Radius
of circle is 5.
b.
i.
If point N(0,-2) lies on the circle then distance of point P from center of the circle must be equal to radius of the circle.
Expression to find distance between two given points and
is:
We have two points with coordinates and
. Therefore;
Hence point N(0,-2) lies on the circle.
ii.
We are required to sketch the circle.
ü Find the radius and coordinates of the center of the circle
From (a) we have found that coordinates of center of the circle are and radius of circle is 5.
ü Indicate the center
ü Mark the four points which show the ends of the horizontal and vertical diameters.
ü Draw the circle to pass through these four points.
ü If any intercepts with the coordinate axes are integers, normally they should also be indicated
iii.
We are required to find an equation of the normal to the circle at the point N(0,-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 N(0,-2) from equation of the radius of the circle CN.
We have coordinates of two points on the radius CN as N(0,-2) and C(-3,2).
Two-Point form of the equation of the line is;
Therefore;
c.
i.
Expression to find distance between two given points and
is:
We have two points with coordinates and
. Therefore;
ii.
Radius is always perpendicular to tangent.
If we have point P(2, 6) outside circle and then a tangent is drawn from point P to the circle, say at point S(x,y), then PS will be always perpendicular to the radius of the circle CS.
This makes a right-triangle CSP with angle CSP=90o. We can apply Pythagoras theorem to the triangle.
Pythagorean Theorem
We know that (from (a:ii)), and
(from C:i). Therefore;
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