$(1)\; 3x+4y-38 = 0$

$(2)\; 3x-4y-38 = 0$

$(3)\; 3x+4y+12 = 0$

$(4)\; 3x-4y-12 = 0$

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- 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.

Given the circle $x^2+y^2-6x-2y-15=0 \rightarrow$ centre $C = (3,1)$

Radius = $\sqrt{3^2+1^2+15} = \sqrt{25} = 5$

We have to find a tangent parallel to $3x+4y+20 = 0$. This will be of the form $3x+4y+k=0$, where $k$ is any constant.

Now, $3x+4y+k=0$ is a tangent to the circle, which means it is perpendicular to radius of length $5$ from $(3,1)$

$\Rightarrow 5 = \large\frac{ \left | 3(3) + 4(1) + k \right |}{\sqrt{3^2+4^2}}$$ = \large\frac{\left | 13+k \right |}{\sqrt{9+16}}$$ = \large \frac{\left| 13+k \right|}{5}$

$\Rightarrow \left|13+k\right| = 25 \rightarrow k = 12$ or $k=-38$

Therefore, the tangents are $3x+4y+12 = 0$ or $3x+4y-38 = 0$

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