Answer:
a). Center of the circle = (-2, 5)
b). Equation of the line ⇒ y = [tex]-\frac{4}{5}x+\frac{58}{5}[/tex]
Step-by-step explanation:
Equation of the circle is,
x² + 4x + y²- 10y + 20 = 30
a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30
[x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30
(x + 2)² + (y - 5)²- 29 + 20 = 30
(x + 2)² + (y - 5)²- 9 = 30
(x + 2)² + (y - 5)² = 39
By comparing this equation with the standard equation of a circle,
Center of the circle is (-2, 5).
b). A point (2, 10) lies on this circle.
Slope of the line joining this point to the center (-2, 5),
[tex]m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
= [tex]\frac{10-5}{2+2}[/tex]
= [tex]\frac{5}{4}[/tex]
Let the slope of the tangent which is perpendicular to this line is '[tex]m_{2}[/tex]'
Then by the property of perpendicular lines,
[tex]m_{1}\times m_{2}=-1[/tex]
[tex]\frac{5}{4}\times m_{2}=-1[/tex]
[tex]m_{2}=-\frac{4}{5}[/tex]
Now the equation of the line passing though (2, 10) having slope [tex]m_{2}=-\frac{4}{5}[/tex]
y - y' = [tex]m_{2}(x-x')[/tex]
y - 10 = [tex]-\frac{4}{5}(x-2)[/tex]
y - 10 = [tex]-\frac{4}{5}x+\frac{8}{5}[/tex]
y = [tex]-\frac{4}{5}x+\frac{8}{5}+10[/tex]
y = [tex]-\frac{4}{5}x+\frac{58}{5}[/tex]
Therefore, equation of the line will be, y = [tex]-\frac{4}{5}x+\frac{58}{5}[/tex]
