1. Pupils notes for Circle Lessons

2.  The equation of a circle with centre ( a, b ) and radius r is We usually leave the equation in this form without multiplying out the brackets SUMMARY  To determine whether a point lies on, inside, or outside a circle, substitute the coordinates of the point into the l.h.s. of the equation of the circle and compare the answer with

3. Since the distance of the point from the centre is less than the radius, the point ( 2, 1 ) is inside the circle e.g. Find the equation of the circle with centre ( 4, -3 ) and radius 5. Does the point ( 2, 1 ) lie on, inside, or outside the circle? Substituting the coordinates ( 2, 1 ): l.h.s. Solution: Using the formula, the circle is

4. To find the centre and radius of a circle given in a form without brackets: Complete the square for the x -terms Complete the square for the y -terms Collect the constants on the r.h.s. Compare with The centre is ( a, b) and the radius is r. SUMMARY

5. e.g. Find the centre and radius of the circle with equation Finally collect the constant terms onto the r.h.s. Solution: we can see the centre is ( 3, 2 ) and the radius is 5. By comparing with the equation,

6. Tangent: No points of intersection: 2 points of intersection: SUMMARY  The discriminant of the quadratic equation formed by eliminating y from the equations of a straight line and a circle tells us how the line and circle are related.

7.  The tangent to a circle is perpendicular to the radius at its point of contact  The perpendicular from the centre to a chord bisects the chord  The angle in a semicircle is a right angle Properties of Circles Diagrams are nearly always needed when solving problems involving circles. A line perpendicular to a tangent to any curve is called a normal. The radius of a circle is therefore a normal.

8. e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3) Solution: Substitute the point that is on the tangent, (5, 7): x (2, 3) (5, 7) x tangent gradient or

9. x chord Centre C is C Solution: The point M (4, 3) is the mid-point of a chord. Find the equation of this chord. e.g. A circle has equation

10. x e.g. A circle has diameter AB where A is ( -1, 1) and B is (3, 3). Show that the point P (0, 0) lies on the circle. diameter A(-1, 1) B(3, 3) Method: If P lies on the circle the lines AP and BP will be perpendicular. Solution: P(0, 0) Hence and P is on the circle. Gradient of AP : Gradient of BP : So,.