If a1=1. And r=1/5 find a8
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Answer:
(x -1)^2 + (y -2)^2 = 13
Step-by-step explanation:
The two given points are the end points of a chord, so the center will be on its perpendicular bisector. That is, the center will be at the point of intersection of the given line and the perpendicular bisector of the given chord.
To find that point, we can write the equation of the perpendicular bisector, then solve the simultaneous equations.
The perpendicular bisector can be written as ...
Δx(x -(3+4)/2) +Δy(y -(5+0)/2) = 0
(3 -4)(x -7/2) + (5 -0)(y -5/2) = 0
-x -9 +5y = 0 . . . . . eliminate parentheses
We can add 3 times this equation to the given equation to find the solution for the circle center.
(3x +2y) + 3(-x -9 +5y) = (7) + 3(0)
17y -27 = 7
y = 34/17 = 2
x = 5y -9 = 1
The circle center is (h, k) = (1, 2).
The square of the radius can be found by substituting one of the given points into the formula for the circle. That formula is ...
(x -h)^2 + (y -k)^2 = r^2
Filling in the values for (h, k) and the first given point, we find ...
(4 -1)^2 + (0 -2)^2 = r^2 = 13
The standard form equation for the circle through the given points with the center on the given line is ...
(x -1)^2 + (y -2)^2 = 13
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<em>Comments on equation for a line</em>
There are a lot of ways to write the equation of a line. Often, I like to use standard form: ax +by = c. When given two points, (x1, y1) and (x2, y2), this can take the form ...
Δy(x -x1) -Δx(y -y1) = 0
where (Δx, Δy) = (x2 -x1, y2 -y1).
The perpendicular line through some point (h, k) will be of the form ...
Δx(x -h) +Δy(y -k) = 0
Note the change in sign for the second term and the switching of Δx and Δy. This is what makes the slope be the negative reciprocal of the slope of the above line through the two points.
For the perpendicular bisector, the point (h, k) needs to be the midpoint of the segment between (x1, y1) and (x2, y2). That midpoint is the average of the two segment endpoints:
(h, k) = ((x1+x2)/2, (y1+y2)/2)
