given the circle x²+y²=10, give the equations of the lines passing through (5,5) and tangent to the circle.?

Question

so we have the circle x²+y²=10, which means its center is at (0,0) and it has a radius of √10. it has two tangents (i think it's two) that pass through the point (5,5). what are the general equations (form ax+bx+c=0) of the two lines?

that's all that's given, it cannot be impossible, and i've seen something like it on Answers. immediate help would be gladly gladly appreciated.

@jcsuperstar714:

(y-5)/(x-5) = -x/y gives y = -x + 2

how??

Answer 1

x^2 + y^2 = 10

differentiating: 2x + 2yy' = 0 or y' = -x/y

That is also the slope of the line through (x,y) and (5,5).

(y-5)/(x-5) = -x/y gives y = -x + 2

x^2 + (-x+2)^2 = 10 --> x^2 -2x -3 = 0 or (x-3)(x+1)=0

This will give you x = 3 and x = -1. x = 3 has y = +/-1 and x = -1 has y = +/-3.

You should be able to make the correct selection of the two points and get the lines. Good luck!

Added to answer the additional question:

(y-5)/(x-5) = -x/y ( = y', the tangent slope)
y(y-5) = -x(x-5)
y^2 - 5y = -x^2 + 5x
x^2 + y^2 = 5x + 5y, but we know x^2 + y^2 = 10
10 = 5(x + y)
2 = x + y
y = -x + 2

Answer 2

Also without differentiation. Since the lines are tangent they form right angles at the point of intersection and hence right triangles. From the Pythagorean theorem the distance from (5,5) squared - the radius squared = distance from (5,5) to the point of tangency.
5²+5²-10=d²=40
If x,y are the coordinates of the point of tangency then the distance formula gives (5-x)²+(5-y)² = 40
or x²-10x+y²-10y = -10 but to be on the circle x²+y² = 10 so
-10x-10y=-20 or x+y=2
so for the circle x²+(-x+2)² = 10
2x²-4x+4 = 10
x²-2x+1 = 4
(x-1)²=4
x=±2+1 or 3,-1
So the points of tangency are (3,±1),(-1,±3) but only (3,-1) and (-1,3) are at the right distance. Using these with the point (5,5) we get equations
m=(5+1)/(5-3) and m = (5-3)/(5+1) so m = -3,1/3
y=-3x+b and y = 1/3x+b
b=20 and b=10/3
y=-3x+20 and y=1/3x+10/3

Answer 3

Differentiate the equation with respective to x,
2x+2yy' = 0
y' = -x/y = -x/[±√(10-x^2)]......(1)
Slope between the tangent point (x, ±√(10-x^2)) and (5,5) is,
y' = (5±√(10-x^2))/(5-x)......(2)
Solve (1) and (2) for x and y,
x = -1, y = 3 => Equation: -x+3y = 10
x = 3, y = -1 =>Equation: 3x-y = 10
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Ideas: When you solve the system of (1) and (2), pay attention to the right branches of y.

Answer 4

First, to find the tangents, find the slopes. its much easier to derive the equation using slopes and y-intercept, which y = mx + b. Note that the tangent lines are PERPENDICULAR to the radius.

Having that known, the relationship of the slopes are m1 * m2 = -1.

To find the slope of the line, m = (y2 - y1) divided by (x2- x1) where m is the slope and x and y are the two coordinates.

use the radius to identify the slope, consider both positive and negative slopes, since its a circle.

once the slope is identified, find its perpendicular equivalent as shown earlier. use that slope and the given coordinate to identify the equation of the tangents.

that's it! good luck, hope this is helpful.

Answer 5

without differentiation
The lines are y-5=m(x-5)
y= 5+m(x-5).Intercept with x^2+y^2=10
x^2+25+10m(x-5) +(x-5)^2-10=0
2x^2 +x(10m-10) +40-50m=0.There must be only one solution(double root) so
(5m-5)^2-4(20-25m)=0
25m^2+50m-55=0
5m^2+10m-11=0 gives you the slopes.
m=((-10+-sqrt(320)/10 = 1+-sqrt(3.2)