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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 $y_1$?

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

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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.

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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?

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  • $\begingroup$ @ Gaurang Tandon, Where and how does it come from? Any elaboration $\endgroup$ – pi-π Dec 24 '17 at 16:33
  • $\begingroup$ @blue_eyed_... To derive it, you need to assume tangent as $L$ being $y=mx+c$. Then, put $\text{perpendicular distance from center of circle to } L=\text{radius of circle}$ to get value of $c$ directly. You will find this derivation in any standard introductory book on coordinate geometry of circles. This concept is taught along with the form $xx_1+yy_1=r^2$ for the tangents. $\endgroup$ – Gaurang Tandon Dec 24 '17 at 16:36
  • $\begingroup$ @blue_eyed_... See the related derivation here $\endgroup$ – Gaurang Tandon Dec 24 '17 at 16:38
  • $\begingroup$ @blue_eyed_... This is another derivation for same result $\endgroup$ – Gaurang Tandon Dec 24 '17 at 16:39
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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

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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 $x^2+y^2=4$

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