SOURCE: analytic geometry
assuming the question is what is the circle equation?
and if (-2,2) is the center of the circle
the equation should look like this: (x+2)^2+(Y-2)^2=R^2
And now only R is needed.
given 2x-5y+4=0 equation of line perpendicular
we can rearange the equation to be y=(2x+4)/5
from that we can see that the slope of the line is 2/5
And from the fact of perpendicular line we can say that the slope
of the radius line is -2/5.
The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular
For that we would calculate the radius line equation and compare it to the equation of line perpendicular
As mentioned earlier the slope of the radious line is -2/5.
So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates
2= - (2/5)*(-2)+b ------> b=2-4/5=1.2
radius equation is y=-(2/5)x+1.2
Now the cross point is calculated by comparing the equations:
-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1
So the cross point is (1/2,1).
The distance between the points is calculated by the following
Formula:
R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=
SQR(7.25)
Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25
SOURCE: Find an equation of the line containing the given
Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=1, y1=0, x2=0, and x1=-6. You should get a=(1-0)/(0-(-6))=1/6
The y-intercept is the y-cordinate for x=0. Its value is 1.
The equation is then y=(x/6) 1.
SOURCE: find an equation of the line containing the given
Calculate
the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5,
x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2
The equation is y=(1/2)x PLUS b, where b is not known yet.
To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).
The point (3,6) lies on the line, so 6=3/2 PLUS b.
Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5
Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2
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