# How to find Equation of a tangent to a circle which is parallel to some line?

given circle: $$x^2 + y^2 - 2x - 4y - 4 = 0,$$

given line which is parallel to the tangent: $$3x - 4y - 1 = 0,$$

I found out the center and radius of the given circle which came out to be $(1,2)$ and $3$ respectively.

but how do i find out the equation of the tangent if i don't know the point where it touches the circle? i know two parallel lines have the same slope. so the tangent and that line should have slope of $3/4$. but that still doesn't give the points to form the equation

• You need to figure out at what point on the circle the tangent has the given slope. What is the slope of the tangent to this circle at an arbitrary point $(x,y)$? – rogerl Nov 14 '16 at 15:40
• Try to use $y=\frac34x+q$ inside the circle equation's and choose the $q$ that make the discriminant $\Delta=0$. – yngabl Nov 14 '16 at 16:11

You've found the radius and centre, then let the tangent line be $3x-4y+c=0$.

The distance of the tangent from the centre is

\begin{align*} \left| \frac{3(1)-4(2)+c}{\sqrt{3^2+4^2}} \right| &= 3 \\ \left( \frac{c-5}{5} \right)^2 &= 3^2 \\ (c-5)^2-15^2 &= 0 \\ (c-20)(c+10) &= 0 \\ c &= 20 \quad \text{or} \quad -10 \end{align*}

The required tangents are:

$$3x-4y+20=0 \quad \text{or} \quad 3x-4y-10=0$$

The slope of the curve in every point of the circle is $\frac{d}{dx}$ (be careful cause you'll have to restrict the domain). Basically, your goal is to find the point where $\frac{d}{dx}$ equals to the slope of the line: it means the point of the circle where the line you're looking for is tangent. When you find that point, you have the slope, so just plug in the formula.

• so what do i have to differentiate to find that point? – Mohit Saxena Nov 14 '16 at 16:34
• @MohitSaxena the circle equation, where y=f(x). – Vitor C Goergen Nov 14 '16 at 21:24

1At first takeout radius of circle and its centre 2 then take out eqn of line parallel to given line 3 apply the length of perpendicular formula on parallel line equation and find value of constant 3length of perpendicular is equal to radius of circle and value of x and y is centre of circle 4keep value of k in parallel line eqn 5 then after keeping value of k that eqn is required eqn of tangent

\begin{array}{ll} \text{Circle:} & (x - 1)^2 + (y - 2)^2 = 9 \\ \text{Line parallel to a tangent:} & 3x - 4y = 1 \end{array}

The line perpendicular to the parallel lines and passing through the center of the circle will pass through the two points where the parallel lines are tangent to the circle.

The perpendicular line must have the form $4x + 3y = c$.

Since it passes through the point $(1,2)$, the center of the circle, then $c = 4(1)+3(2) = 10$.

So the equation of the perpendicular line is $4x+3y= 10$

We find the two points where this line intersects the circle.

\begin{align} 4x+3y &= 10 \\ y &= -\frac 43x + \frac{10}{3} \\ y-2 &= -\frac 43x + \frac 43 \\ y-2 &= -\frac 43(x -1) \\ (y-2)^2 &= \frac{16}{9}(x-1)^2 \\ \hline (x - 1)^2 + (y - 2)^2 &= 9 \\ (x - 1)^2 + \frac{16}{9}(x-1)^2 &= 9 \\ (x-1)^2 &= \frac{81}{25} \\ x &= 1 \pm \frac 95 \\ (x,y) &\in \left\{ \left(\frac{14}{5}, -\frac 25 \right), \left(-\frac 45, \frac{22}{5} \right) \right\} \end{align}

The form of the parallel lines is $3x-4y = c$

For the first point $c = 3\left(\frac{14}{5} \right) - 4\left(-\frac 25 \right) = 10$.

For the second point $c = 3\left(-\frac 45 \right) - 4\left(\frac{22}{5} \right) = -20$.

So the equations of the two lines is $3x-4y=10$ and $3x-4y=-20$.