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CIRCLE

CIRCLE

IN THIS LESSON WE ARE GOING TO STUDY ABOUT EQUATION OF CIRCLE GENERAL EQUATION OF CIRCLE TANGENT AND NORMAL CONDITION OF TANGENCY AND SOME EXAMPLES WITH EXERCISES TO CONCLUDE

THE CIRCLE

SO WHAT IS A CIRCLE? IN SIMPLE LANGUAGE, A circle is a shape consisting of all points in a plane which are equidistance to the given fixed point i.e. Centre. And the distance between fixed point and the curved plane is radius. Diameter is the straight line passing through Centre and joining two any points in a circle.

Figure: showing a circle

Formulae

Equation of a circle: If the Centre of the circle is at A( h,k ) and radius is ‘a’ then the equation of the circle is given by (?-h) 2 + (?- k) 2 = a 2 Note ; I f the Centre is at origin then h=k=0. Then the equation of circle be ? 2 + ? 2 = a 2

What is a tangent? T angent is a straight line which just touches the curve at a given point. In this figure ATB is tangent at T. The point T is called the point of contact. NOTE: The tangent should only touch the only one point in a circle.

Normal Normal is a straight line which is perpendicular to the tangent . In this figure, The straight line BOC is the normal to the Tangent AB.

A

B

O

C

Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y – axis .

Solution: Since the circle has intercept ‘a’ from x – axis, the circle must pass through (a, 0) and (-a, 0) as it already passes through the origin. Since the circle has intercept ‘b’ from x – axis, the circle must pass through (0, b) and (0, -b) as it already passes through the origin. Let us assume the circle passing through the points A(a,0) and B(0,b). We know that the equation of the circle with AB as diameter is given by (x – x 1 ) (x – x 2 ) + (y – y 1 ) (y – y 2 ) = 0 (x – a) (x – 0) + (y – 0) (y – b) = 0 x 2 + y 2 + ax + by = 0 or x 2 + y 2 – ax – by = 0 ∴The equation of the circle is x 2 + y 2 + ax + by = 0 or x 2 + y 2 – ax – by = 0

Find the equation of the circle which circumscribes the triangle formed by the line 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0

Given: The lines 2x + y – 3 = 0 x + y – 1 = 0 3x + 2y – 5 = 0 On solving these lines we get the intersection points A(2, – 1), B(3, – 2), C(1,1) So by using the standard form of the equation of the circle: x 2 + y 2 + 2ax + 2by + c = 0….. (1) Substitute the points (2, -1) in equation (1), we get 2 2 + (- 1) 2 + 2a(2) + 2b(- 1) + c = 0 4 + 1 + 4a – 2b + c = 0 4a – 2b + c + 5 = 0….. (2) Substitute the points (3, -2) in equation (1), we get

3 2 + (- 2) 2 + 2a(3) + 2b(- 2) + c = 0 9 + 4 + 6a – 4b + c = 0 6a – 4b + c + 13 = 0….. (3) Substitute the points (1, 1) in equation (1), we get 1 2 + 1 2 + 2a(1) + 2b(1) + c = 0 1 + 1 + 2a + 2b + c = 0 2a + 2b + c + 2 = 0….. (4) Upon simplifying equations (2), (3), (4) we get a = -13/2, b = -5/2, c = 16 Now by substituting the values of a, b, c in equation (1), we get x 2 + y 2 + 2 (-13/2)x + 2 (-5/2)y + 16 = 0 x 2 + y 2 – 13x – 5y + 16 = 0 ∴ The equation of the circle is x 2 + y 2 – 13x – 5y + 16 = 0

Show that the points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle, and find its equation, Centre, and radius.

Solution: Given: The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle. Let us assume the circle passes through the points A, B, C. So by using the standard form of the equation of the circle: x 2 + y 2 + 2ax + 2by + c = 0….. (1) Substituting A (5, 5) in (1), we get, 5 2 + 5 2 + 2a(5) + 2b(5) + c = 0 25 + 25 + 10a + 10b + c = 0 10a + 10b + c + 50 = 0….. (2) Substitute the points B (6, 4) in equation (1), we get, 6 2 + 4 2 + 2a(6) + 2b(4) + c = 0 36 + 16 + 12a + 8b + c = 0 12a + 8b + c + 52 = 0….. (3)

Substitute the point C (-2, 4) in equation (1), we get, (-2) 2 + 4 2 + 2a(-2) + 2b(4) + c = 0 4 + 16 – 4a + 8b + c = 0 20 – 4a + 8b + c = 0 4a – 8b – c – 20 = 0….. (4) Upon simplifying equations (2), (3) and (4) we get, a = – 2, b = – 1 and c = – 20 Now by substituting the values of a, b, c in equation (1), we get x 2 + y 2 + 2(- 2)x + 2(- 1)y – 20 = 0 x 2 + y 2 – 4x – 2y – 20 = 0 ….. (5) Substituting D (7, 1) in equation (5) we get, 7 2 + 1 2 – 4(7) – 2(1) – 20 49 + 1 – 28 – 2 – 20 0 ∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) lie on a circle.

Now let us find the centre and the radius. We know that for a circle x 2 + y 2 + 2ax + 2by + c = 0, Centre = (-a, -b) Radius = √ (a 2 + b 2 – c) Comparing equation (5) with equation (1), we get Centre = [-(-4)/2, -(-2)/2 )]= (2, 1) Radius = √ (2 2 + 1 2 – (-20)) = √ (25) = 5 ∴ The centre and radius of the circle is (2, 1) and 5.

Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x – 2y + 4 = 0 . Find the equation of the circle whose centre lies on the positive direction of y – axis at a distance 6 from the origin and whose radius is 4 . If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle . A circle whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 passes through the origin. Find its equation . Find the equation of the circle which passes through (3, – 2), (- 2, 0) and has its centre on the line 2x – y = 3. Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are con – cyclic . Find the equation of the circle which circumscribes the triangle formed by the line x + y + 3 = 0, x – y + 1 = 0 and x = 3 Find the equation of the circle the end points of whose diameter are the centres of the circles x 2 + y 2 + 6x – 14y – 1 = 0 and x 2 + y 2 – 4x + 10y – 2 = 0 .

Thank you hope you understood well!! ? Prepared by D eewash Sharma