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# Which of the following are tangents to the circle $x^2+y^2-4x+8y+12=0$ from $(5-3)$?

$(1)\; x+y+2=0$

$(2)\; x+y-2= 0$

$(3)\; 7x-y+32= 0$

$(4)\; 7x+y-32=0$

Toolbox:
• The general equation of the circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$, where the centre is $(-g, -f)$
• The radius of the circle is $\sqrt{g^2+f^2-c}$
• The perpendicular distance from centre of circle to the tangent is equal to the radius.
• We can find the slopes of the equation $y-y_1 =m(x-x_1)$ and using that, we can write down the equation for the two tangents.
Given the circle$x^2+y^2-4x+8y+12=0 \rightarrow$ centre $C = (2,-4)$
Radius = $\sqrt{2^2+-4^2-12} = \sqrt{8} = 2 \sqrt 2$
We have a point $(5,-3)$, so the equation of the tangents $= y - (-3) = m (x - 5) \rightarrow mx - y + (-3 - 5m) = 0$
Now, this is a tangent to the circle, which means it is perpendicular to radius of length $2\sqrt 2$ from $(2,-4)$
$\Rightarrow 2\sqrt 2 = \large\frac{ \left | m(2) -1 (-4) + (-3-5m) \right |}{\sqrt{m^2+-1^2}}$$= \large\frac{\left | 2m+4-5m-3 \right |}{\sqrt{m^2+1}}$$ = \large \frac{\left| -3m+1 \right|}{\sqrt{m^2+1}}$
$\Rightarrow$ Squaring both sides, $(2\sqrt2)^2 \times \sqrt{m^2+1}^2 = (-3m+1)^2 \rightarrow 8m^2+8 = 9m^2-6m+1$
$\Rightarrow m^2-6m-7 = 0 \rightarrow (m-7)(m+1) = 0$
$\Rightarrow m = 7$ or $m=-1$
Case 1: $m = 7$:
Equation of the tangent $= y-(-3) = 7(x+5) \rightarrow 7x-y + 32 = 0$
Case 2: $m = -1$:
Equation of the tangent $= y-(-3) = -1(x+5) \rightarrow x+y-2 = 0$
edited Apr 9, 2014