Complete the square on the y-terms by adding 4 to both sides.



Factor the trinomial:



Compare to the standard form of a circle:



And we see that the center of the circle is (h,k) = (2,-2)
and the radius is r=2

So we sketch the figure:



Let X and Q be the points of tangency, we draw radii OX and OQ,
then we draw OP and draw XY parallel to the x-axis.

We want the equations of PX and PQ.

We now see that the equation of the tangent PQ 
is the vertical line x = 4. 

We also see that PQ is 9 units in length, 7 units above the x-axis
and 2 units below the x-axis.

From right triangle OPQ, we see that





We use the identity 

 

Since , by right triangle PXY,

 and  are complementary and
 

Finally we use the point-slope formula to find the equation of tangent PX













That's the equation of the tangent PX

and the equation of the tangent PQ is the vertical line x = 4.

Edwin

You could probably also find the slanted tangent line by 

the equation of the circle and the point (4,7) with

y = mx + b

and finding the values of b and m that would produce
a double root, by setting the discriminant = 0.

Edwin

Yes it can be done that way:

We use the point-slope formula for the equation of a line through (4,7) 




Substitute in





That simplifies to the quadratic in x2



The point of tangency can be thought of as where 
the tangent line intersects the circle TWICE at
the same point.  So both solutions must be equal.

So we set the  



That simplifies to 





The equation of the slanted tangent line is





Which simplifies to what I got above.

Edwin