Answer:
[tex]y+2=\dfrac{4}{7}(x-1)[/tex]
Step-by-step explanation:
We can implicitly differentiate the circle equation to find the general form of its derivative:
[tex]\dfrac{d}{dx}(x^2 + y^2 + 2x- 3y+13)= \dfrac{d}{dx}(0)[/tex]
↓ applying the power and chain rules
[tex]2x + 2y \cdot \dfrac{dy}{dx} + 2 - 3y \cdot \dfrac{dy}{dx} + 0 = 0[/tex]
↓ isolating the dy/dx's on one side
[tex]2y \cdot \dfrac{dy}{dx} - 3 \cdot \dfrac{dy}{dx} = -2x-2[/tex]
↓ factoring out the dy/dx's
[tex]\dfrac{dy}{dx} \cdot (2y - 3) = -2x-2[/tex]
↓ dividing both sides by (2y - 3)
[tex]\dfrac{dy}{dx} = \dfrac{-2x-2}{2y-3}[/tex]
Now, we can find the slope of the tangent line to the circle at (1, -2) by plugging those coordinates into the general form of the derivative:
[tex]\left \dfrac{dy}{dx}\right|_{x=1,\ y=-2} = \dfrac{-2(1)-2}{2(-2)-3}[/tex]
[tex]= \dfrac{-4}{-7}[/tex]
[tex]= \dfrac{4}{7}[/tex]
Finally, we can plug this slope into the point-slope form equation, inputting the coordinates again:
[tex]y-b=m(x-a)[/tex]
where [tex]m=\text{slope}[/tex], [tex](a,b)=\text{a point on the line}[/tex].
↓↓↓
[tex]y-(-2)=\dfrac{4}{7}(x-1)[/tex]
[tex]\boxed{y+2=\dfrac{4}{7}(x-1)}[/tex]
Further Note
Re-arranging the circle's equation into standard form, we get:
[tex](x^2+2x+1) +(y^2- 3y+9/4)=-13-1-9/4[/tex]
[tex](x+1)^2 +(y- \frac{3}{2})^2=-\frac{65}{4}[/tex]
Since the right side (radius squared) is negative, this circle doesn't actually exist. Therefore, the hypothetical tangent line we calculated doesn't have any real significance on the Cartesian plane.