I know the formula for equation of tangents for a standard circle $x^2+y^2=a^2$ which is $$y=mx\pm a\sqrt{1+m^2}$$ From where we can find the slope of the tangents. Is there any such equation for a circle of the form $x^2+y^2+2gx+2fy+c=0$
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1All you have to do is shift the origin to $(-g,-f)$ and replace $a$ with $\sqrt{g^2+f^2-c}$. – Paras Khosla Jan 26 '20 at 10:44
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@ParasKhosla so the equation will be $y-f=(x-g)m\pm \sqrt{(g^2+f^2-c)(1+m^2)}$? – Aditya Jan 26 '20 at 10:57
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It will be $(y+f)=m(x+g)\pm r\sqrt{1+m^2}$. – Paras Khosla Jan 26 '20 at 11:32
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@ParasKhosla why would it be +? Isn’t $x=X+h$ where h is the new origin? – Aditya Jan 26 '20 at 11:39
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@Aditya it wouldn't be $+$ because the standard form of a circle with centre $(-g, -f)$ is $(x-(-g))^2 +(y-(-f))^2=a^2$ where $a$ is the radius. – Tapi Jan 27 '20 at 16:23
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For all real values (including infinite) of $m,$ the straight line will be tangent to the circle.
Hint:
For any conic in 2D $$ax^2+2hxy+by^2+2gx+2fy+c=0$$,
let $y=mx+d$ be the tangent.
Replace the value of $y$ in terms of $x$ to form a quadratic equation in $x$.
Each root will represent the abscissa of intersection.
For tangency, the two abscissas must coincide, i.e., the roots must be same
lab bhattacharjee
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See also : https://math.stackexchange.com/questions/930926/equation-of-circle-tangent-to-line-with-radius – lab bhattacharjee Jan 26 '20 at 10:47
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In the standard equation of tangent to circle $y=mx\pm a\sqrt{1+m^2}$. Replace $(x,y)$ with $(x+g,y+f)$ and radius $a$ with $r=\sqrt{g^2+f^2-c}$. $$. $$ Hence equation of tangent to circle $x^2+y^2+2gx+2fy+c=0$ is $$y+f=m(x+g) \pm r\sqrt{1+m^2}$$
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