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) |
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Method
1
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Substitute
points in the general form of circle and solve coefficients
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Let the circle be (2,1) : (0,5) : (-1,2) : |
Start with the General Form of circle. Substitution of
P, Q, R. |
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2D + E + F = -5 ….. (1) 5E + F = -25 ………(2) D – 2E – F = 5 …….(3) |
Simplification |
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(1) + (3), 3D – E = 0 ……. (4) (2) + (3), D + 3E = -20 ….. (5) From (4), E = 3D …………(6) Subst. (6) in (5), D + 9D = -20 \ D = -2 ………………. (7) Subst. (7) in (6), E = -6 …….. (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. |
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The equation of the circle
is
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Substitute D,
E, F in the General
Form. |
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Find the equation of the circle passing through the points
P(2,1), Q(0,5), R(-1,2) |
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Method
2
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Use Centre and Radius Form of the circle.
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Let the center and radius
of the circle be C(a,b) and r. |PC| = |QC| = |RC|
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The distance from center to the given 3 points are equal. |
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5 - 4a - 2b = 25 - 10b = 5
+ 2a - 4b
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Simplification. Note that there
are no “square” terms. Get a simultaneous
equation in a and b. |
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(2) – (1), 5b = 15 \ b = 3 ….. (3) Subst. (3) in (1), a = 1 \ C = (3, 1) |
Solve a and b. The center of the circle can be found. |
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r = |PC| |
Solve for the
radius. |
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The equation of the circle
is |
Use the Center- radius form to get the equation of
circle. |
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Find the equation of the circle passing through the points
P(2,1), Q(0,5), R(-1,2) |
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Method
3
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The perpendicular bisectors of two chords meet at the
centre.
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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. |
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(1) – (2), \ x = 1 …. (3) Subst. (3) in (1), \ y = 3 \ C = (1,3) |
Solve for the
center. |
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r = |PC| |
Solve for the
radius. |
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The equation of the circle
is |
Use the Center- radius Form to get the equation of
circle. |
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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) |
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Method
4
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Converse of Angle in semi-circle.
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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 |
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Equation of L1:
Equation of L2:
(1) – (2), \ x = 0 ….. (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. |
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P = (-6, 5) S = (0, -3) PS is the diameter of the
circle. The equation of the
circle: (x + 6)(x - 0) + (y – 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. |
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Find the equation of the circle passing through the points
P(-6,5), Q(-3,-4), R(2,1) |
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Method
5
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System of circles.
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Equation of PQ :
L : 3x + y +
13 = 0 Circle with PQ as
diameter: C : (x +6)(x+3) + (y –5)(y
+ 4) = 0
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Two points
form. Write PQ in general form. Use the
diameter form to find the circle with PQ as diameter. |
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The system of circle passing
through the intersections of the circle C and the line L can be given by
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This system of circles must pass through points P and Q. |
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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 :
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Find k. The system of circles passes through P, Q. The circle also passes through R. |
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Find the equation of the circle passing through the points
P(-6,5), Q(-3,-4), R(2,1) |
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Method
6
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Angles in same segment .
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Gradient of PQ Gradient of PR Let a be the angle QPR.
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You may investigate the
case for which the absolute value is added:
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Let S(x, y) be any point
on the circle. Gradient of SQ Gradient of SR Let b be the angle QSR.
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We do not
include the absolute value in the formula. The order of m
is therefore important. |
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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°-b tan a = - tan b the negative
sign is taken care. |
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Find the equation of the circle passing through the points
P(-6,5), Q(-3,-4), R(2,1) |
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Method
7
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Determinant method
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The required circle is of
the form:
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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. |
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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. |
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