A circle has centre on line x+y=5 and passes through points (6,6) and (5,7). Find equation of secant parallel to given line at distance 7/sqrt2 units.
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The center of the circle lies on the line y=-x+5. Two points on the circle are (6,6) and (5,7).
(1) We can find the center in a couple of ways. The distance from the center to each of the given points is the same. Let the center be (x,5-x). (Since it lies on the line y=5-x)
Then use the distance formula from this point to each of the given points and set them equal:
Square both sides and expand:
The center of the circle is at (2,3).
** We could also use the fact that the center lies on the perpendicular bisector of the segment from (5,7) to (6,6) and also on the line y=-x+5 to find the center. We would find the intersection of the lines y=-x+5 and y=x+1 to be (2,3) as before **
(2) Let (x,y) be the point on the secant line that is also on the segment drawn perpendicular to the secant from the center of the circle. Since (x,y) lies on the segment perpendicular to the secant line, the point can be written as (x,x+1).
** The slope from the point to the center will be 1 as it is perpendicular to the given line which has slope -1. Then `(y-3)/(x-2)=1 ==>y=x+1` **
Since the distance from this point to the center is given as `d=7/sqrt(2)` we can again apply the distance formula:
Squaring both sides and expanding we get:
** Note that we get two answers; this is because there are two possible secant lines fitting the description.**
(3) If x=-1.5 then y=-.5. The slope of the secant line is -1 so the equation of the secant line is y=-x-2
If x=5.5 then y=6.5. Again the slope is -1, so the equation is y=-x+12
(4) The graphs:
The radius of the circle is 5. The distance to the secant is `7/sqrt(2)~~4.95`
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