Answer:
The equation of the line is [tex]y = -x-3[/tex].
Step-by-step explanation:
Let be [tex]x^{2}+y^{2}-4\cdot x +10\cdot y +4 = 0[/tex] and [tex]x^{2}+y^{2}+10\cdot x +4\cdot y + 25 = 0[/tex] the general equations of the two circles, which must be transformed into their standard forms, which are:
[tex](x-h)^{2}+(y-k)^{2} = r^{2}[/tex]
Where:
[tex]h[/tex], [tex]k[/tex] - Components of the center of the circle, dimensionless.
[tex]r[/tex] - Radius of the circle, dimensionless.
1) [tex]x^{2}+y^{2}-4\cdot x +10\cdot y +4 = 0[/tex] [tex]\wedge[/tex] [tex]x^{2}+y^{2}+10\cdot x +4\cdot y + 25 = 0[/tex] Given
2) [tex](x^{2}-4\cdot x + 4) + (y^{2}+10\cdot y +25) = 25\,\wedge \,(x^{2}+10\cdot x +25)+(y^{2}+4\cdot y + 4) = 4[/tex] Associative property/Compatibility with addition
3) [tex](x-2)^{2}+(y+5)^{2} = 25\,\wedge \,(x+5)^{2}+(y-2)^{2} = 4[/tex] Perfect square trinomial/Result
The centers of both circles are [tex](2,-5)[/tex] and [tex](-5, 2)[/tex], respectively.
A line is represented by the following expression:
[tex]y-y_{o} = m\cdot (x-x_{o})[/tex]
Where:
[tex]m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
If [tex](x_{1}, y_{1}) = (2,-5)[/tex], [tex](x_{2},y_{2}) = (-5,2)[/tex] and [tex](x_{o},y_{o}) = (2,-5)[/tex], the equation of the line is:
[tex]m = \frac{2-(-5)}{-5-2}[/tex]
[tex]m = -1[/tex]
[tex]y - (-5) = -1 \cdot (x-2)[/tex]
[tex]y + 5 = -x +2[/tex]
[tex]y = -x-3[/tex]
The equation of the line is [tex]y = -x-3[/tex].
