In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:

where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained. 

  which is the slope of the tangent line.

To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.

, the perpendicular slope can then be calculated as 

To find the slope of the tangent line, we must take the derivative.

By using the Power Rule we will be able to find the derivative:

Therefore derivative of  is .

Now we plug in , giving us .

The only two bits of information that are given for the tangent line is that it runs through the points  and . With these two points, the line's slope can be easily calculated through the equation: 

where  is slope,  is the -coordinate of the points, and  is the -coordinates of the points. 

Slope can be calculated through substituting in for the given values:

The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by  

The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.

The equation of the line is y – 4 = (3/4)(x – (–3))

Rearranging gives us: 3x – 4y = -25

The graph of the equation  is a circle with center .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints  and  will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope. 

The tangent line therefore has equation

Rewrite the equation of the circle in standard form to find its center:

Complete the square:

The center is . 

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints  and  will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope. 

The tangent line therefore has equation

To find an equation tangent to

we need to find the first derviative of this equation with respect to  to get the slope  of the tangent line.

So,

due to power rule .

First we need to find our slope by plugging our  into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point  we plug the point into

.

Therefore our equation becomes,

Once we rearrange, the equation is

To find an equation tangent to

we need to find the first derviative of this equation with respect to  to get the slope  of the tangent line.

So,

due to power rule .

First we need to find the slope by plugging our  into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point  we plug it into

.

Therefore our equation is

Once we rearrange, the equation is

To find an equation tangent to

we need to find the first derviative of this equation with respect to  to get the slope  of the tangent line.

So,

due to power rule .

First we need to find the slope by plugging in our  into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point  we plug our point into

.

Therefore our equation is

Once we rearrange, the equation is

To find an equation tangent to

we need to find the first derviative of this equation with respect to  to get the slope  of the tangent line.

So,

due to power rule .

First we need to find our slope by plugging in our  into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point  we plug the point into

.

Therefore our equation is

Once we rearrange, the equation is