Question

X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square

Answer 1

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = $-\frac{4}{5}x+\frac{58}{5}$

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),

    $m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

          = $\frac{10-5}{2+2}$

          = $\frac{5}{4}$

    Let the slope of the tangent which is perpendicular to this line is '$m_{2}$'

    Then by the property of perpendicular lines,

          $m_{1}\times m_{2}=-1$

          $\frac{5}{4}\times m_{2}=-1$

                 $m_{2}=-\frac{4}{5}$

   Now the equation of the line passing though (2, 10) having slope $m_{2}=-\frac{4}{5}$

           y - y' = $m_{2}(x-x')$

           y - 10 = $-\frac{4}{5}(x-2)$

           y - 10 = $-\frac{4}{5}x+\frac{8}{5}$

                  y = $-\frac{4}{5}x+\frac{8}{5}+10$

                  y = $-\frac{4}{5}x+\frac{58}{5}$

Therefore, equation of the line will be, y = $-\frac{4}{5}x+\frac{58}{5}$