Tangent lines and circles...?

Question

A tangent line to a circle is a line that intersects the circle at exactly one point. The tangent line in perpendicular to the radius of the circle at this point of contact. Write the equation of a line tangent to the circle whose equation is x^2 + y^2 = 25 at the point (3,-4).
Okay, so I know that you have to use the slope formula and y=mx+b, but I just don't know how...Please help me!!

Answer 1

First of all, there is a mistake in rfamilyme's solution.
The slope of the tangent line should be 3/4, not -3/4.
But we don't even need calculus here.
The centre of the circle is at the origin (0,0).
Draw the radius to the point (3,-4). The
slope of the line joining these two points is -4/3.
Since the tangent line is perpendicular to the
radius at this point its slope is the negative
reciprocal of -4/3 or 3/4.
So the answer is y + 4 = 3/4(x-3)
or y = 3/4 x - 25/4.

Answer 2

in order to locate the equation of a line, you want the slope and some extent that you recognize is on the line. The slope is straightforward: a tangent to a circle is perpendicular to the radius on the point the position the line will be tangent to the circle. In different words, the radius of your circle starts at (0,0) and is going to (3,4). This slope is an identical as upward push/run = (y2-y1)/(x2-x1) = (4-0)/(3-0) = 4/3. A line it somewhat is perpendicular to this radius line may have a slope it somewhat is the unfavourable reciprocal of the slope you recognize. In different words, the line we favor may have a slope of -3/4. the point all of us understand is on the line is (3,4). So the usual equation of the line is: (y-y0) = m(x-x0), the position (x0, y0) is the point all of us understand is on the line, or (3,4) for that reason m is the slope, -3/4 for that reason x,y stay as variables (y - 4) = (-3/4)(x - 3) y - 4 = (-3/4)x + 9/4 y = (-3/4)x + 25/4

Answer 3

differentiating
2x+2y dy/dx=0
2y dy/dx=-2x
dy/dx=-x/y so at(3,-4) it is -3/4
equation of the tangent
y-(-4)=-3/4(x-3)
4y+16=-3x+9
3x+4y+7=0