2010
02.08
02.08
1) A circle has the following properties:
a) The x-axis and the line y=20 are tangents to the circle
b) The circle passes through the points (0,2) and (0,8)
c) The centre lies in the first quadrant.
Find the equation of the circle
I have found the radius to be 10, but i can’t find the centre. Any help?
radius = 10
let the center be ( h, k)
the equatipon of the circle with center (h, k) and radius 10 is
( x-h)^2 + ( y-k)^2 = 10^2
the circle passes through ( 0,2 )
so x=0, y=2 –> h^2 +(2-k)^2=100
h^2 +k^2- 4k +4 -100 =0
h^2 +k^2- 4k - 96 =0
passes through (0,8) also, so x=0, y=8
we have h^2 +(8-k )^2 =100
h^2 +k^2 - 16k +64 -100 =0
h^2 +k^2 - 16k - 36 =0
h^2 +k^2- 4k - 96 =0 —-(1)
h^2 +k^2 - 16k - 36 =0—(2)
(1) - (2) —> 12k - 60 =0
k = 60/12 =5
h^2 +(2-k)^2=100, k=5
h^2 +(-3)^2 =100
h^2 = 100 - 9
= 91
h = root (91)
center is ( root 91, 5), in first quadrant
so the equation is ( x- root91) ^2 + (y-5) ^2 = 100
simplifying,
x^2 +y^2 -2 (root91) x -10y +16 =0
You are off to a good start. Given that the circle passes through the two points specified, why not write expressions for the center point (x, y) using the distance formula and the known radius. Both of the points given must be 10 units from the common center.
10 = sqrt(x^2 + (y - 2)^2) and 10 = sqrt(x^2 + (y - 8)^2) if you square both sides of these and set the right sides equal (since both equal 100), you get:
x^2 + y^2 -4y + 4 = x^2 + y^2 -16y + 64 which can be simplified to:
-4y + 4 = -16y + 64
12y = 60
y = 5 (This presents a bit of a problem, since given point a above, the y coordinate should be 10. The line connecting the points of tangency must be a diameter, and the center would lie at its midpoint. )
Use either original equation to find x.
x^2 + 5^2 - 4*5 + 4 = 100
x^2 + 25 - 20 + 4 = 100
x^2 = 91, so x = sqrt(91) We may ignore the negative root, since we have been told the centre is in the first quadrant.
There is some inconsistency in this information. If you draw a circle based upon b, you will see that the center must be at y = 5, but drawing one based upon a, it must be at y = 10.
Footnote: While either condition a or condition b may be satisfied by an infinite number of possible circles, there is no solution that satisfies both conditions, with or without the addition of condition c. However, if condition a is modified to say the circle is tangent to the line y = 10 and the x-axis, there is a unique (and very tidy) solution with radius 5 and its center at (4, 5). I wonder if perhaps that was the original problem? If so, the answer would be:
(x - 4)^2 + (y - 5)^2 = 25
the question must be wrong because the points (0,8) and (0,2) are forming a chord and according to my leckie notes the centre of the circle lies on the perpindicular bisector of a chord. the perpindicular bisector of this chord would meen that the radius must be 5. but you are not wrong saying it is 10 because it should also be half of y=20, so the question is either giving wrong co-ordinates or a wrong equation.
general equation of a circle
(x-h)^2+(y-k)^2=r^2 where (h,k) is centre
you know k=10 as y=20 and y=0 are tangents
therefore (x-h)^2+(y-10)^2=100
at point (0,2)
(0-h)^2+(2-10)^2=100
(0-h)^2+64=100
(0-h)^2=36
h=6
therefore centre of circle is (6,10)
I did get y=5 initially but the other information is inconsistent with this, i haven’t checked whether the point (0,8) will lie on this circle, although i suspect not!, think you might need to check the question
(I’m sorry my englishis not so good i”m from a french spoken country)
the general equation of the cercle is (x-a)^2+(y-b)^2=R^2 in which the centre is (a,b) et the radius R.
As you already have the radius. Let use this equation by putting in it the coordinate of the two points.
With (0,2) we have this a^2 + (2-b)^2 = R^2
With (0,8) we have this a^2+ (8-b)^2 = R^2
We thus have a^2 + (2-b)^2 = a^2+ (8-b)^2
(2-b)^2 = (8-b)^2
This equation gives us two equation when putting out the squar.
2 - b = 8 - b
or
2 - b = - (8-b)
Only the second has a solution: 2 - b = -8 + b
10 = 2 * b
So b=2
We remplace b=2 , R=10 in this equation a^2 + (2-b)^2 = R^2 to find out a.
We have a^2+ (2-2)^2= 10^2
a^2=10^2
Resolving this give a=10 or a= -10
Only the positive answer is taken as it is the first quadrant.
The center therefore is (10 , 2).
From a/ —->radius = 10, y -center = 10
but form b/ —>y-center = (2+8)/2 = 5 !!!! Can not do