Find the point where $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$ Find the coordinates of the point where the line $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$.
Can someone please show me the first few steps of solving the simultaneous equation?
 A: The clever method: use geometry
We know from the question that the line in question is tangent to the circle.  A tangent line is perpendicular to the radius from the center of the circle to the point of tangency.  That is, the point of intersection between the line in circle must also intersect the line of perpendicular slope through the origin (which is the center of our circle).  Since the line we're given is $y = \frac14(3x+25)$, we know that the tangent line has slope $3/4$, which means that the desired radius has slope $-4/3$.  Since the radius passes through the origin, its equation is
$$
y = -\frac 43 x
$$
Now, to find the intersection of the radius with the circle, we make the above substitution for $y$, giving us
$$
x^2 + \left(-\frac 43 x\right)^2 = 25
$$
Solving this gives you two solutions for $x$, one of which is the one we need (hint: in which quadrant do the lines intersect?).  Once you have $x$, use our nifty substitution to find $y$.
Alternatively, you could substitute $y$ in the equation for the other line, giving you
$$
3x - 4\left(-\frac 43 x\right) + 25 = 0
$$
Which you could use to solve for $x$.
A: Solve the first equation for $y$, and substitute into the second equation
$$
3x - 4y + 25 = 0\\
4y = 3x + 25\\
y = \frac14(3x+25)
$$
Now, substitute into the second equation to get:
$$
x^2 + \left(\frac14(3x+25)\right)^2 = 25
$$
Can you take it from there?
A: First up:
Equation of the line is : $3x-4y+25=0$
This line will cross the x-axis at F $(-25/3,0)$ and y-axis at E $(0,25/4)$, and also will be tangent to this given circle at point G.

The Line EF is perpendicular to AG at G.
Length of AG=5, AF=25/3 units. Using pythagoras theorem in $\triangle AGF$ will yield coordinate for G. You can do the math!
PS: Geogebra is quite time consuming tool. I spent half hour just to draw this picture. :(
A: $$
3x - 4y + 25 = 0\\
4y = 3x + 25\\
m = \frac34
$$
Differentiate circle equation $ x^2 + y^2 =25 $
$ \dfrac{dy}{dx}= -\dfrac xy = m = \dfrac 34 $
Solve for x and y to get,  $ x = \mp 3, y =\pm 4 $
Only the first set $ (-3,4)$ satisfies the given equation of straight line tangent.
