Find the equation of the tangents to the circle

Question

Find the equation of the tangents to the circle $x^2+y^2=4$ which are parallel to $3x+4y-5=0$

My Attempt: The equation of tangent to the circle $x^2+y^2=4$ at point $P(x_1,y_1)$ is given by $$xx_1+yy_1=4$$ How do I get $x_1$ and $y_1$?

Answer 1

The equation of any straight line parallel to $$3x+4y-5=0$$ is $$3x+4y-c=0$$ where $c$ is an arbitrary constant.

Now it will be tangent to the given circle if the radius = the distance of $3x+4y-c=0$ from the center

i.e., $$2=\left|\dfrac{3\cdot0+4\cdot0-c}{\sqrt{3^2+4^2}}\right|\iff c=?$$

Answer 2

There are more than one forms of the tangents to a circle, the key is to realize when to use which form of tangent. Here, you're given the slope of the tangent, hence, the best way to proceed is to use the slope form of the tangent to a circle:

$$y=mx\pm r\cdot\sqrt{1+m^2}$$

Can you proceed now?

Answer 3

By C-S $$4=x^2+y^2=\frac{1}{25}(3^2+4^2)(x^2+y^2)\geq\frac{1}{25}(3x+4y)^2.$$ The equality occurs, when $$3x+4y=10$$ and $$3x+4y=-10$$ are touching to the circle.

Answer 4

Given two lines having equation $r:ax+by+c=0;\;r':a'x+b'y+c'=0$

Then $r\parallel r' \Leftrightarrow ab'-a'b=0$

In this case

$xx_1+yy_1-4=0$ is parallel to $3x+4y-5=0$ if

$\begin{cases} 4x_1-3y_1=0\\ x_1^2+y_1^2=4\\ \end{cases}$

Which gives

$$\left(\frac{6}{5};\;\frac{8}{5}\right);\;\left(-\frac{6}{5};\;-\frac{8}{5}\right)$$

and then

$3 x+4 y-10=0;\;3 x+4 y+10=0$

Hope this helps

Answer 5

From where you have left off,

$$xx_1+yy_1=4\ \ \ \ (1)$$ will be parallel to $$3x+4y-5=0\ \ \ \ (2)$$ iff $\dfrac{x_1}3=\dfrac{y_1}4=F$(say)

$(1)$ becomes $$3Fx+4Fy=4\ \ \ \ (2)$$

Again using algebraic rule,

$$F=\pm\sqrt{\dfrac{x_1^2+y^2_1}{3^2+4^2}}=?$$ as $(x_1,y_1)$ lies on $x^2+y^2=4$