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?
[Math] Find the point where $3x-4y+25=0$ is a tangent to the circle $x^2+y^2=25$
algebra-precalculus
Related Solutions
"Passing through" means that it is on the circle i.e. it is one of the points on the circle.
How to solve:
You are given a point on the circle, $(-2, 1)$, a tangent line $3x - 2y = 6$ and a tangent point $(4, 3)$.
Find the slope of the tangent line:
$3x - 2y = 6$
$-2y = -3x + 6$
$y = \frac{3}{2}x + 3$
The slope is $\frac{3}{2}$.
Recall that the slope of a perpendicular line is the negative reciprocal of the original line. In other words, if $m_1$ is the slope of a line, and $m_2$ is the slope of the perpendicular line, then $m_1 * m_2 = -1$.
The radius must be perpendicular to the tangent line. Using the above formula, let $m_1$ be equal to the slope of the tangent line, $\frac{3}{2}$. Then $m_2$, the slope of the radius, can be found like so:
$\frac{3}{2} * m_2 = -1$
$m_2 = -\cfrac{1}{\left(\cfrac{3}{2}\right)}$
$m_2 = -1 * \frac{2}{3}$
$m_2 = -\frac{2}{3}$
Using point-slope form, derive an equation of a line that the centre is on.
$y - y_0 = m(x - x_0)$
$y - 3 = -\frac{2}{3}(x - 4)$
$y = -\frac{2}{3}x - \frac{8}{3} + 3$
$y = -\frac{2}{3} + \frac{1}{3}$
Let the coordinates of the center be $(a, b)$. Then the equation of the circle is $(x-a)^2 + (y-b)^2 = r^2$
Remember that $b = -\frac{2}{3}a + \frac{1}{3}$
Substitute $b$ for the expression above, and plug in the point $(-2, 1)$ into the equation.
$(-2-a)^2 + (1-(-\frac{2}{3}a + \frac{1}{3}))^2 = r^2$
$a^2 + 4a + 4 + (\frac{2}{3}a - \frac{2}{3})^2 = r^2$
$a^2 + 4a + 4 + \frac{4}{9}a^2 - \frac{8}{9}a + \frac{4}{9} = r^2$
$\frac{13}{9}a^2 + \frac{28}{9}a + \frac{40}{9} = r^2$
Do the same thing, but instead, plug in the point $(4, 3)$ into the equation.
To make this faster, $(4-a)^2 + (3-(\frac{2}{3}a + \frac{1}{3}))^2$ evaluates to $a^2 - 8a + 16 + (-\frac{2}{3}a + \frac{8}{3})^2$, which equals $a^2 - 8a + 16 + \frac{4}{9}a^2 - \frac{32}{9}a + \frac{64}{9}$. Further simplification yields $\frac{13}{9}a^2 - \frac{40}{9}a + \frac{80}{9}$.
Now you have two equations in two variables. This creates a system.
$\left\{ \begin{array}{rcl} \frac{13}{9}a^2 + \frac{28}{9}a + \frac{40}{9} = r^2 \\ \frac{13}{9}a^2 - \frac{40}{9}a + \frac{80}{9} = r^2 \\ \end{array} \right.$
Multiply both equations by $9$ to get rid of the fractions. This allows for easier work.
$\left\{ \begin{array}{align} 13a^2 + 28a + 40 = 9r^2 \\ 13a^2 - 9a + 80 = 9r^2 \\ \end{array} \right.$
Equate the expressions and solve.
$13a^2 + 28a + 40 = 13a^2 - 9a + 80$
$37a - 40 = 0$
$37a = 40$
$a = \frac{40}{37}$
Plug in the value of $a$ into the equation $b = -\frac{2}{3}a + \frac{1}{3}$ defined above.
$b = -\frac{2}{3} * \frac{40}{37} + \frac{1}{3}$
$b = -\frac{80}{111} + \frac{1}{3}$
$b = -\frac{80}{111} + \frac{37}{111}$
$b = -\frac{57}{111}$
The center of the circle is at $(\frac{40}{37}, -\frac{57}{111})$.
To find $r^2$, use the distance formula, but square both sides so that the square root is removed. Allow me to demonstrate:
Original distance formula: $D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, where $D$ is the radius $r$, and $(x_1, y_1)$ and $(x_2, y_2)$ are any two given points.
Squaring both sides gives $(x_1 - x_2)^2 + (y_1 - y_2)^2 = r^2$, which looks a lot like the equation of a circle (it actually is not).
Plug in the points $(\frac{40}{37}, -\frac{57}{111})$ and $(-2, 1)$ into the formula.
$(\frac{40}{37} + 2)^2 + (-\frac{57}{111} - 1)^2 = r^2$
$(\frac{114}{37})^2 + (-\frac{168}{111})^2 = r^2$
$\frac{12996}{1369} + \frac{28224}{12321} = r^2$
$\frac{198762372}{16867449} = r^2$
The equation of the circle is:
$(x - \frac{40}{37})^2 + (y + \frac{57}{111})^2 = \frac{198762372}{16867449}$
I might have made a calculation mistake / error, but you get the idea of how to do it. I hope this answer helped you!
EDIT: Yes I did make a calculation error when I was solving the system, but you get the point. DO NOT use my answer for your assignment / homework, because it is wrong.
Since you know the center and the point of tangency, you can compute the slope of the radius to the point of tangency. Since the center is $(3, 0)$ and the point of tangency is $(2, 2\sqrt{2})$, the slope of the radius to the point of tangency is
$$m_r = \frac{2\sqrt{2} - 0}{3 - 2} = 2\sqrt{2}$$
The slope of the tangent line to the circle is perpendicular to the radius at the point of tangency. If two non-vertical lines are perpendicular, their slopes are negative reciprocals, so the slope of the tangent line to the circle at $(2, 2\sqrt{2})$ is the negative reciprocal of the slope of the radius to the point of tangency. Thus, the tangent line has slope
$$m_{\perp} = -\frac{1}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{4}$$
You can then use the point-slope equation
$$y - y_0 = m(x - x_0)$$
to write the equation of the tangent line, where $(x_0, y_0)$ is the point $(2, 2\sqrt{2})$ and $m$ is the slope of the tangent line. The equation of the tangent line to the circle $(x - 3)^2 + y^2 = 9$ at $(2, 2\sqrt{2})$ is
$$y - 2\sqrt{2} = -\frac{\sqrt{2}}{4}(x - 2)$$
A reflection in the $x$-axis sends point $(x, y)$ to the point $(x, -y)$. Thus, the reflection of the point $(2, 2\sqrt{2})$ in the $x$-axis is $(2, -2\sqrt{2})$. To find the equation of the tangent line to the circle at this point, follow the steps outlined above with $(2, -2\sqrt{2})$ replacing $(2, 2\sqrt{2})$.
Best Answer
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 $x$.