Find the Center and Radius Then Write the Standard Equation of Each Circle Brainly

Equation of circle passing through 3 given points

Find the equation of the circle passing through the points P(2,1), Q(0,5), R(-1,2)
Method 1	Substitute points in the general form of circle and solve coefficients
Let the circle be (2,1) : (0,5) : (-1,2) :		Start with the General Form of circle. Substitution of P, Q, R.
2D + E + F = -5 �K.. (1) 5E + F = -25 �K�K�K(2) D �V 2E �V F = 5 �K�K.(3)		Simplification
(1) + (3), 3D �V E = 0 �K�K. (4) (2) + (3), D + 3E = -20 �K.. (5) From (4), E = 3D �K�K�K�K(6) Subst. (6) in (5), D + 9D = -20 \ D = -2 �K�K�K�K�K�K. (7) Subst. (7) in (6), E = -6 �K�K.. (8) Subst. (7), (8) in (1), F = -5		Solve the equations. Eliminating F and get a simultaneous equation in D and E. Solve D and E and finally get F.
The equation of the circle is		Substitute D, E, F in the General Form.

Find the equation of the circle passing through the points P(2,1), Q(0,5), R(-1,2)
Method 2	Use Centre and Radius Form of the circle.
Let the center and radius of the circle be C(a,b) and r. |PC| = |QC| = |RC|		The distance from center to the given 3 points are equal.
5 - 4a - 2b = 25 - 10b = 5 + 2a - 4b		Simplification. Note that there are no ��square�� terms. Get a simultaneous equation in a and b.
(2) �V (1), 5b = 15 \ b = 3 �K.. (3) Subst. (3) in (1), a = 1 \ C = (3, 1)		Solve a and b. The center of the circle can be found.
r = |PC|		Solve for the radius.
The equation of the circle is		Use the Center- radius form to get the equation of circle.

Find the equation of the circle passing through the points P(2,1), Q(0,5), R(-1,2)
Method 3	The perpendicular bisectors of two chords meet at the centre.
Let L1 and L2 be the perpendicular bisectors of PQ and QR respectively. Mid-point of PQ Mid-point of QR Gradient of PQ Gradient of QR Since L1 ^ PQ, L2 ^ QR, Gradient of L1 Gradient of L2 = L1 : L2 :		Find the mid-points of two chords. Find the gradient of two chords. Product of gradients of perpendicular lines = -1 Find the equations of perpendicular bisectors using gradient-point form.
(1) �V (2), \ x = 1 �K. (3) Subst. (3) in (1), \ y = 3 \ C = (1,3)		Solve for the center.
r = |PC|		Solve for the radius.
The equation of the circle is		Use the Center- radius Form to get the equation of circle.

We use another set of three points to work for the other methods below.

Find the equation of the circle passing through the points P(-6,5), Q(-3,-4), R(2,1)
Method 4	Converse of Angle in semi-circle.
Gradient of PQ Gradient of PR Let L1 ^ PQ and L2 ^ PR. Let L1 and L2 meet at S. Gradient of L1 = Gradient of L2 = 2		Product of gradients of perpendicular lines = -1
Equation of L1: Equation of L2: (1) �V (2), \ x = 0 �K.. (3) Subst. (3) in (1), y = -3. \ S = (0, -3)		Use Point-gradient form to find equations of L1 and L2. Solve the equations, the point S can be found.
P = (-6, 5) S = (0, -3) PS is the diameter of the circle. The equation of the circle: (x + 6)(x - 0) + (y �V 5)(y + 3) = 0		Apply the theorem of Converse of Angle in Semi-circle we can see that P, Q, R, S are concyclic with PS as diameter. Use diameter form to find equation of circle.

Find the equation of the circle passing through the points P(-6,5), Q(-3,-4), R(2,1)
Method 5	System of circles.
Equation of PQ : L : 3x + y + 13 = 0 Circle with PQ as diameter: C : (x +6)(x+3) + (y �V5)(y + 4) = 0		Two points form. Write PQ in general form. Use the diameter form to find the circle with PQ as diameter.
The system of circle passing through the intersections of the circle C and the line L can be given by		This system of circles must pass through points P and Q.
We like to find one of the circles in this system which passes through the point R (2,1). Subst. R(2,1) in C + kL we have, \ k = -1 The required circle is :		Find k. The system of circles passes through P, Q. The circle also passes through R.

Find the equation of the circle passing through the points P(-6,5), Q(-3,-4), R(2,1)
Method 6	Angles in same segment .
Gradient of PQ Gradient of PR Let a be the angle QPR. �K..(1)		You may investigate the case for which the absolute value is added:
Let S(x, y) be any point on the circle. Gradient of SQ Gradient of SR Let b be the angle QSR. �K.. (2)		We do not include the absolute value in the formula. The order of m is therefore important.
Since a = b , therefore tan a = tan b Equate (1) and (2), we get Simplify, we get the equation of circle:		Use the Angles in the same segment . If we consider the absolute value previously, we have to add �� sign in (1) and (2). Use Opposite angles of cyclic quad. supplementary , we get a = 180 �X - b tan a = - tan b the negative sign is taken care.

Find the equation of the circle passing through the points P(-6,5), Q(-3,-4), R(2,1)
Method 7	Determinant method
The required circle is of the form:		Reasons: (1) The equation is of second degree and is a conic. (2) Coefficients of x^2 = y^2 and no xy- term, it is a circle. (3) Subst. P(-6, 5) in (*), we get R1 and R2 are equal and therefore P satisfies (*). Similarly for Q and R. Therefore the circle passes through P, Q, R.
Using Laplace expansion we get: Simplify, we get the equation of circle:		Although (*) is very simple and compact, its evaluation is complicate. You should know how to break down a higher order determinant.

Find the Center and Radius Then Write the Standard Equation of Each Circle Brainly

Source: https://qc.edu.hk/math/Advanced%20Level/circle%20given%203%20points.htm

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