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Published byKevin Lindsey Modified over 9 years ago
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Pupils notes for Circle Lessons
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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
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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
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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
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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,
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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.
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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.
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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
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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
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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,.
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