Centre and radius
A point P ( x , y ) P(x, y) P ( x , y ) ( a , b ) (a, b) ( a , b ) r r r r r r
x y (a, b) r P A circle (x − a)² + (y − b)² = r²
( x − a ) 2 + ( y − b ) 2 = r 2 (x - a)^2 + (y - b)^2 = r^2 ( x − a ) 2 + ( y − b ) 2 = r 2 Expanding gives the general form x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2 + y^2 + 2gx + 2fy + c = 0 x 2 + y 2 + 2 g x + 2 f y + c = 0 ( − g , − f ) (-g, -f) ( − g , − f ) g 2 + f 2 − c \sqrt{g^2 + f^2 - c} g 2 + f 2 − c x 2 x^2 x 2 y 2 y^2 y 2
Worked example · NECO 2023
NECO 2023 · Paper 2 · Q2 (a)
Find the equation of the circle whose centre is ( 4 , − 5 ) (4, -5) ( 4 , − 5 ) ( − 3 , 2 ) (-3, 2) ( − 3 , 2 )
The radius
r 2 = ( 4 + 3 ) 2 + ( − 5 − 2 ) 2 = 49 + 49 = 98 {r^2 = (4 + 3)^2 + (-5 - 2)^2 = 49 + 49 = 98} r 2 = ( 4 + 3 ) 2 + ( − 5 − 2 ) 2 = 49 + 49 = 98 Think first. r is the distance from (4, −5) to (−3, 2).
The equation
( x − 4 ) 2 + ( y + 5 ) 2 = 98 {(x - 4)^2 + (y + 5)^2 = 98} ( x − 4 ) 2 + ( y + 5 ) 2 = 98 Expand: x 2 − 8 x + 16 + y 2 + 10 y + 25 = 98 {x^2 - 8x + 16 + y^2 + 10y + 25 = 98} x 2 − 8 x + 16 + y 2 + 10 y + 25 = 98
So x 2 + y 2 − 8 x + 10 y − 57 = 0 {x^2 + y^2 - 8x + 10y - 57 = 0} x 2 + y 2 − 8 x + 10 y − 57 = 0
← Back Next step → Show all steps Start again
Check yourself
The circle with centre ( 3 , − 1 ) (3, -1) ( 3 , − 1 ) x 2 + y 2 − 6 x + 2 y + c = 0 x^2 + y^2 - 6x + 2y + c = 0 x 2 + y 2 − 6 x + 2 y + c = 0 c c c
More: centre and radius
NECO 2023 · Paper 1 · Q29 Find the equation of the circle which has its centre at ( − 2 , 4 ) (-2, 4) ( 1 , 5 ) (1, 5) WAEC 2017 · Paper 2 · Q10 If f ( x ) = 2 x − 3 ( x 2 − 1 ) ( x + 2 ) f(x) = \dfrac{2x - 3}{(x^2 - 1)(x + 2)} x x f ( x ) f(x) f ( x ) f(x) WAEC 2016 · Paper 2 · Q11 Without using mathematical tables or a calculator, solve 2 + log 10 x − log 10 20 = log 10 ( x 2 + 4 ) 2 + \log_{10} x - \log_{10} 20 = \log_{10}(x^2 + 4) WAEC 2018 · Paper 2 · Q4 The radius of the circle x 2 + y 2 − 4 x − 2 y + C = 0 x^2 + y^2 - 4x - 2y + C = 0 3 2 3\sqrt2 WAEC 2022 · Paper 2 · Q11 Write down the matrix M M T : ( x , y ) → ( 3 x + y , − 2 x + 3 y ) T : (x, y) \to (3x + y, -2x + 3y) WAEC 2023 · Paper 2 · Q4 An equation of a circle is 3 x 2 + 3 y 2 + 6 x + 2 m y − 18 = 0 3x^2 + 3y^2 + 6x + 2my - 18 = 0 m m 10 \sqrt{10} WAEC 2008 · Paper 2 · Q9 The sum of the first n n S n = 5 n 2 2 + 5 n 2 S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2} WAEC 2008 · Paper 2 · Q10 The equation of a circle is x 2 + y 2 − 4 x + 2 y + c = 0 x^2 + y^2 - 4x + 2y + c = 0 c c 2 3 2\sqrt3 WAEC 2020 · Paper 1 · Q24 A circle with centre ( 5 , − 4 ) (5, -4) ( 5 , 0 ) (5, 0) WAEC 2020 · Paper 1 · Q27 Which of the following is not an equation of a circle? WAEC 2022 · Paper 1 · Q33 The equation of a circle is given as 2 x 2 + 2 y 2 − x − 3 y − 41 = 0 2x^2 + 2y^2 - x - 3y - 41 = 0 WAEC 2023 · Paper 1 · Q33 Find the radius of the circle 2 x 2 + 2 y 2 − 4 x + 5 y + 1 = 0 2x^2 + 2y^2 - 4x + 5y + 1 = 0 Tangents
The tangent at a point P P P P P P − 1 ÷ -1 \div − 1 ÷
x y (a, b) r P tangent The tangent at P gradient of tangent = −1 ÷ gradient of radius Centre, radius and tangent Set g, f and c, then move P
−7 −5 −3 −1 1 3 5 7 −6 −4 −2 2 4 6 x y P (1, −0.5) centre (−g, −f) 2.5 radius −1.19 tangent gradient at P
g = −1 f = 0.5 c = −5 P (angle round the circle) = 40°
x² + y² − 2x + 1y − 5 = 0: centre (1, −0.5), r² = g² + f² − c = 1 + 0.25 + 5 = 6.25, so r = 2.5. At P(2.92, 1.11) the radius has gradient 0.84, so the tangent has gradient −1.19. Worked example · NECO 2023
NECO 2023 · Paper 2 · Q9 (a, b)
Find the centre and radius of the circle 3 x 2 + 3 y 2 + 12 x − 6 y − 45 = 0 3x^2 + 3y^2 + 12x - 6y - 45 = 0 3 x 2 + 3 y 2 + 12 x − 6 y − 45 = 0
Find the equation of the tangent to the circle at ( 2 , 3 ) (2, 3) ( 2 , 3 ) y y y x x x
Centre and radius
x 2 + y 2 + 4 x − 2 y − 15 = 0 {x^2 + y^2 + 4x - 2y - 15 = 0} x 2 + y 2 + 4 x − 2 y − 15 = 0 g = 2 {g = 2} g = 2 f = − 1 {f = -1} f = − 1 The centre is ( − 2 , 1 ) {(-2, 1)} ( − 2 , 1 ) r = 4 + 1 + 15 = 20 = 2 5 {r = \sqrt{4 + 1 + 15} = \sqrt{20} = 2\sqrt5} r = 4 + 1 + 15 = 20 = 2 5
Think first. Divide by 3 first, so x² has coefficient 1.
The radius to (2, 3)
Its gradient is 3 − 1 2 + 2 = 1 2 {\frac{3 - 1}{2 + 2} = \frac12} 2 + 2 3 − 1 = 2 1
The tangent
Its gradient is − 2 {-2} − 2 y − 3 = − 2 ( x − 2 ) {y - 3 = -2(x - 2)} y − 3 = − 2 ( x − 2 )
So y = − 2 x + 7 {y = -2x + 7} y = − 2 x + 7
← Back Next step → Show all steps Start again
Check yourself
The point ( 5 , 1 ) (5, 1) ( 5 , 1 ) x 2 + y 2 − 4 x + 6 y − 12 = 0 x^2 + y^2 - 4x + 6y - 12 = 0 x 2 + y 2 − 4 x + 6 y − 12 = 0 y y y x x x
A circle through three points
Put each point into x 2 + y 2 + 2 g x + 2 f y + c = 0 x^2 + y^2 + 2gx + 2fy + c = 0 x 2 + y 2 + 2 g x + 2 f y + c = 0 g g g f f f c c c
Worked example · WAEC 2018
WAEC 2018 · Paper 2 · Q9
A circle is drawn through the points ( 3 , 2 ) (3, 2) ( 3 , 2 ) ( − 1 , − 2 ) (-1, -2) ( − 1 , − 2 ) ( 5 , − 4 ) (5, -4) ( 5 , − 4 )
coordinates of the centre of the circle;
Three equations
( 3 , 2 ) {(3, 2)} ( 3 , 2 ) 6 g + 4 f + c = − 13 {6g + 4f + c = -13} 6 g + 4 f + c = − 13 ( − 1 , − 2 ) {(-1, -2)} ( − 1 , − 2 ) − 2 g − 4 f + c = − 5 {-2g - 4f + c = -5} − 2 g − 4 f + c = − 5 ( 5 , − 4 ) {(5, -4)} ( 5 , − 4 ) 10 g − 8 f + c = − 41 {10g - 8f + c = -41} 10 g − 8 f + c = − 41 Think first. Substitute (3, 2), (−1, −2) and (5, −4).
Remove c
First minus second: 8 g + 8 f = − 8 {8g + 8f = -8} 8 g + 8 f = − 8 g + f = − 1 {g + f = -1} g + f = − 1
Third minus first: 4 g − 12 f = − 28 {4g - 12f = -28} 4 g − 12 f = − 28 g − 3 f = − 7 {g - 3f = -7} g − 3 f = − 7
Subtract: 4 f = 6 {4f = 6} 4 f = 6 f = 3 2 {f = \frac32} f = 2 3 g = − 5 2 {g = -\frac52} g = − 2 5
Think first. Subtract the equations in pairs.
c, the centre and the radius
c = − 13 − 6 ( − 5 2 ) − 4 ( 3 2 ) = − 13 + 15 − 6 = − 4 {c = -13 - 6\left(-\frac52\right) - 4\left(\frac32\right) = -13 + 15 - 6 = -4} c = − 13 − 6 ( − 2 5 ) − 4 ( 2 3 ) = − 13 + 15 − 6 = − 4 The centre is ( − g , − f ) = ( 5 2 , − 3 2 ) {(-g, -f) = \left(\frac52, -\frac32\right)} ( − g , − f ) = ( 2 5 , − 2 3 )
r = 25 4 + 9 4 + 4 = 12.5 ≈ 3.536 {r = \sqrt{\frac{25}{4} + \frac94 + 4} = \sqrt{12.5} \approx 3.536} r = 4 25 + 4 9 + 4 = 12.5 ≈ 3.536 The equation is x 2 + y 2 − 5 x + 3 y − 4 = 0 {x^2 + y^2 - 5x + 3y - 4 = 0} x 2 + y 2 − 5 x + 3 y − 4 = 0
← Back Next step → Show all steps Start again
Where a line meets a circle
Make x x x y y y
x y M N A line cutting a circle Substitute the line into the circle and solve
Worked example · WAEC 2013
WAEC 2013 · Paper 2 · Q4
The line 2 y = x + 3 2y = x + 3 2 y = x + 3 x 2 + y 2 − 2 x + 6 y − 15 = 0 x^2 + y^2 - 2x + 6y - 15 = 0 x 2 + y 2 − 2 x + 6 y − 15 = 0 M M M N N N N N N M M M N N N
Substitute
( 2 y − 3 ) 2 + y 2 − 2 ( 2 y − 3 ) + 6 y − 15 = 0 {(2y - 3)^2 + y^2 - 2(2y - 3) + 6y - 15 = 0} ( 2 y − 3 ) 2 + y 2 − 2 ( 2 y − 3 ) + 6 y − 15 = 0 Expand: 4 y 2 − 12 y + 9 + y 2 {4y^2 - 12y + 9 + y^2} 4 y 2 − 12 y + 9 + y 2 − 4 y + 6 + 6 y − 15 = 0 {- 4y + 6 + 6y - 15 = 0} − 4 y + 6 + 6 y − 15 = 0
5 y 2 − 10 y = 0 {5y^2 - 10y = 0} 5 y 2 − 10 y = 0 5 y ( y − 2 ) = 0 {5y(y - 2) = 0} 5 y ( y − 2 ) = 0 Think first. From the line, x = 2y − 3.
The points
y = 0 {y = 0} y = 0 x = − 3 {x = -3} x = − 3 y = 2 {y = 2} y = 2 x = 1 {x = 1} x = 1 N N N N ( 1 , 2 ) {N(1, 2)} N ( 1 , 2 ) M ( − 3 , 0 ) {M(-3, 0)} M ( − 3 , 0 ) Think first. y = 0 or y = 2. Which point is in the first quadrant?
← Back Next step → Show all steps Start again
Your turn
(a) Find the equation of the circle centre ( 2 , 3 ) (2, 3) ( 2 , 3 ) y y y 3 x − 2 y + 6 = 0 3x - 2y + 6 = 0 3 x − 2 y + 6 = 0
Show the answer x 2 + y 2 − 4 x − 6 y + 9 = 0 x^2 + y^2 - 4x - 6y + 9 = 0 x 2 + y 2 − 4 x − 6 y + 9 = 0
Worked solution (try it first) The
y y y -intercept is where
x = 0 x = 0 x = 0 :
− 2 y + 6 = 0 -2y + 6 = 0 − 2 y + 6 = 0 , so
y = 3 y = 3 y = 3 : the point
( 0 , 3 ) (0, 3) ( 0 , 3 ) .
r 2 = ( 0 − 2 ) 2 + ( 3 − 3 ) 2 = 4 r^2 = (0 - 2)^2 + (3 - 3)^2 = 4 r 2 = ( 0 − 2 ) 2 + ( 3 − 3 ) 2 = 4 .
( x − 2 ) 2 + ( y − 3 ) 2 = 4 (x - 2)^2 + (y - 3)^2 = 4 ( x − 2 ) 2 + ( y − 3 ) 2 = 4 .
Expand:
x 2 − 4 x + 4 + y 2 − 6 y + 9 = 4 x^2 - 4x + 4 + y^2 - 6y + 9 = 4 x 2 − 4 x + 4 + y 2 − 6 y + 9 = 4 , so
x 2 + y 2 − 4 x − 6 y + 9 = 0 x^2 + y^2 - 4x - 6y + 9 = 0 x 2 + y 2 − 4 x − 6 y + 9 = 0 .
Watch out
The y y y x = 0 x = 0 x = 0 Report a problem with this question