Tangent of a circle help!?

Question

What is the equation of the line tangent to point P on this circle:
(x-2)^2 + y^2=10
P=(1,-3)

Explanation a plus!

Answer 1

Equation of a circle:
(x-h)² + (y-k)² = r²
Center: (h,k)

Center of the circle is (2,0)
Slope of radius to (1, -3) = (0+3)/(2-1) = 3

The tangent must be perpendicular to the radius.
The product of slopes of 2 perpendicular lines would be -1.
Slope of tagent = -1/3

y = mx + c
y = -3, m = -1/3, x = 1, c = ?
-3 = -1/3 + c
c = -3 + 1/3 = -8/3

y = -x/3 - 8/3
3y = -x - 8
x + 3y + 8 = 0

Equation of tangent: x + 3y + 8 = 0

Answer 2

This can be solved using calculus (derivatives) but this seems like an algebra/analytical geometry level problem.

First off, you know that the tangent line contains the point (1,-3). If you can find the slope of it, you're all set. Here's how you find the slope. Given that it's a tangent line, it's perpendicular to that particular radius containing (1,-3).

The center of the circle is at (2,0). The slope of that radius is (-3-0)/(1-2) = -3/-1 = 3. The slope of the tangent is therefore -1/3.

The line has a slope of -1/3 and contains the point (1,-3). It's equation is therefore y+3 = -1/3*(x-1).

Answer 3

The form of the equation for a line is:
y = m*x + b

The line tangent to the circle must have the same slope as the function at that point. To find the slope at that point you must take the derivative of the function with respect to either X or Y, and then plug in the respective value of X or Y into the function. This will give you the slope (m) of the tangent line at that point. Then you need to find the intercept (b) by plugging in the X and Y coordinates of the point you are interested in along with the slope that we just calculated. Now solve for b.

After doing that math you should come up with this equation:
y = -1/3 * x - 8/3

Answer 4

From your eqn of a circle your center point for your circle is:
(x - h)2 + (y - k)2 = r2
Where h and k are the center points, therefore
(-2 , 0)

Now you have two points for a line. You have to work out the gradient between the two points.
SO m=(y2-y2)/(x2-x1) = 3

Now for tangental lines m*m equals = 1
therefore the gradient of the trangent is =1/3

y-b = m(x-a)
where p=(a, b)
So
y+3=1/3(x-1)
moving the three over :- y= 1/3x -8/3

Answer 5

(x-2)^2 + y^2 = 10

The slope of the tangent is derivative of equation of circle

differentiate implicitly

2(x-2) + 2y y' = 0

divide by 2

x - 2 + yy' = 0

yy' = 2-x

y' = (2-x)/y

substituting the given coordinates of P

slope m = (2-1)(-3)

= -1/3

the equation of tangent

y - y1 = m(x-x1)

y +3 = -1/3(x - 1)

multiply with 3

3y + 9 = -x + 1


3y + x + 8 = 0

Answer 6

If x is the attitude between r and AB then: r*cos x = R*cos(one hundred twenty-x) => tan x = (2r+R)/(R*sqrt(3)) Then, section a would be: a = 2*r*cos x = r*R*sqrt(3)/sqrt(R^2+Rr+r^2) **************** Sorry, i replace into calculated it as they're externally tangent, now I see the word "internally"... :)

Answer 7

Take the derivative so you get 2*(x-2)+2y*dy/dx=0

dy/dx = -[2*(x-2)]/(2y)] or (2-x)/y

So the slope of the line is (2-1)/-3 = -1/3

and the point-slope form is y-yo=m*(x-xo)

Where xo-1 and y0=-3

m is the slope or dy/dx

Hope that helps. If you're in calculus.

Answer 8

Differentiate implicitly getting:
2(x-2) +2yy' = 0
2yy' = -2(x-2)
y' = -(x-2)/y
At (1,-3), y'= -(1-2)/-3 = -1/3
So y = -x/3+b
-3 = -1/3 +b --> b= -2 2/3
So y = -x/3 -2 2/3