i need help with this math problem
how do u find the equation of the straight tanget to the circumference on the indicated point.....
x^2 + y^2 =10 .... point is (6,7)
2007-11-03 14:55:01 · 5 answers · asked by X_nOmAd_oo57-ha 3 in Science & Mathematics ➔ Mathematics
point is (3,1) sorry for any inconvenience
2007-11-03 14:55:46 · update #1
Center of the circle is (0,0)
so a line passing through (3,1) has a slope of 1/3
so the tangent is -3 Line is y = -3x + b
1 = -9 + b = 10
so the eq is y = -3x + 10
2007-11-03 15:03:46 · answer #1 · answered by norman 7 · 0⤊ 0⤋
x=2 y=3
2007-11-03 22:04:27 · answer #2 · answered by Jess 4 · 0⤊ 0⤋
The center is (2,2) and the point is (6,7).
First find the slope of the line that passes through these two points using the equation m=(y1-y2)/(x1-x2).
Find the slope of the tangent line by finding the slope the line perpendicular to this line. (1/-m)
Put the new slope into the point-slope form of the equation. (y-y1)=m(x-x1) where (x1,y1) is a point on the line use (6,7). This is the equation of the tangent line.
2007-11-03 22:14:17 · answer #3 · answered by ss89 2 · 0⤊ 0⤋
x^2 + y^2 =10
2x + 2y * dy/dx = 0
dy/dx = -x/y
at point (3,1)
dy/dx = -3
it's the slope of the tangent line.
Suppose the equation of the tangent is y = -3x + b
at (3, 1), 1 = -9 + b, b = 10
so
y = -3x + 10
2007-11-03 23:23:07 · answer #4 · answered by zsm28 5 · 0⤊ 0⤋
WHAT? you just confused me!
2007-11-03 22:00:01 · answer #5 · answered by Anonymous · 0⤊ 0⤋