Linear equation with circle

Question

What is the equation of a line that is tangent to the circle with center at the origin and a radius of 1 and passes through the point through the point (0,2)?

I tried it for quite a while but still can't find a straightforward method of doing this problem. The answer is y=-rad3*X+2

hamam_Abdallah · Answer 1 · 2017-06-01T01:45:13.187

1

Let $ax+by+c=0$ be the equation of the line.

the distance from the center to this tangent line is the radius.

$$\frac {|c|}{\sqrt {a^2+b^2}}=1$$ the point $(0,2) $ belongs to this line, so

$$2b+c=0\;\;,\;\;c=-2b$$

from this

$$a^2+b^2=4b^2\;\;,\;\;a=\pm b \sqrt {3}$$

the equation is then

$$\pm x\sqrt {3}+y=2$$

it is normal to find two lines since the point $(0,2) $ is in a symetry axis.

edited Jun 01 '17 at 01:45

answered Jun 01 '17 at 01:32

hamam_Abdallah

62,951

score 1 · Accepted Answer · answered Jun 01 '17 at 01:47

1

Let $A(0,2)$, $C$ be a tangency point and $O$ be an origin.

Hence, $\measuredangle ACO=90^{\circ}$ and since $OC=\frac{1}{2}AO$, we get $\measuredangle OAC=30^{\circ}$.

Thus, $m_{AC}=\pm\sqrt3$ and we get the answer: $$y-2=\pm\sqrt3x.$$ Done!

answered Jun 01 '17 at 01:47

Michael Rozenberg

194,933

Because the slope of $AC$ it's $\pm\tan\left(90^{\circ}-30^{\circ}\right)=\pm\sqrt3$. – Michael Rozenberg Jun 01 '17 at 02:20
$\frac{1}{\sqrt3}$ There is a better explaining. Let $AC\cap axis- x={D}$. Thus, $m_{AC}=\tan\measuredangle ADx$. – Michael Rozenberg Jun 01 '17 at 02:53
It's intersection $AC$ with $x$-axis. – Michael Rozenberg Jun 01 '17 at 04:38
$AC\cup x-axis={D}$ means $AC$ intersects the $x$- axis in the point $D$. $\measuredangle ADx$ it's an angle between the ray $DA$ and the ray $Dx$, which is a positive direction of the $x$-axis. $m_{AC}$ it's a slope of the line $AC$. Draw it! It's obvious. – Michael Rozenberg Jun 02 '17 at 05:50

Doug M · Answer 3 · 2017-06-01T01:58:35.903

0

Here is a figure from trigonometry that is worth noting.

if $\csc t = 2$ then $t = \frac {\pi}{6}$ and $\sec t = \frac {2}{\sqrt 3}$

Intercept-intercept equation of a line.

$\frac {x}{a} + \frac {y}{b} = 1$

and with these intercepts.

$\frac {x \sqrt 3}{2} + \frac {y}{2} = 1$

And of course it can be reflected across the y axis.

$\frac {x \sqrt 3}{2} - \frac {y}{2} = 1$

edited Jun 01 '17 at 01:58

answered Jun 01 '17 at 01:50

Doug M

57,877

I was wondering why my results were flipped 90 degrees from the others. Fixed – Doug M Jun 01 '17 at 01:55
@Salahamam_Fatima yes, that is what I said. – Doug M Jun 01 '17 at 02:02

Linear equation with circle

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