Question

find the center and radius of the circle x²+y²-6x+4y-36=0​

Answer 1

Answer:

The general equation of a circle is given as:

⇒ x² + y² + 2gx + 2fy + c = 0

According to this equation,

• x-coordinate of center = -g

• y-coordinate of center = -f

• Radius of the circle = √ ( g² + f² - c )

The given equation in the question is:

• x² + y² - 6x + 4y - 36 = 0    ...(i)

Comparing (i) with the general equation we get:

⇒ 2g = -6

⇒ g = -6/2 = -3

⇒ -g = -(-3) = 3

Similarly,

⇒ 2f  = 4

⇒ f = 4/2 = 2

⇒ -f = -2

Radius of the circle is:

⇒ r = √ ( g² + f² - c )

⇒ r = √ [ 9 + 4 - (- 36 )]

⇒ r = √ 49 = 7 units

Hence the center of the circle is ( 3, -2 ) and radius of the circle is 7 units.

Answer 2

$\bf{ \underline{ \underline{Given :- }}}$

$• \: {x}^{2} + {y}^{2} - 6x + 4y - 36 = 0$

$\sf In \: the \: equation,$

$• \: \sf x-coordinate \: of \: center = -g$

$• \: \sf y - coordinate \: of \: center = -f$

$• \: \sf radius \: of \: the \: circle = \sqrt{ {g}^{2} + {f}^{2} - c }$

$\sf According \: to \: the \: given \: equation,$

$\sf - 2g = - 6$

$\sf \rightarrow - g = - 3$

$\sf \rightarrow g = 3$

$\sf \: And ,$

$\sf - 2f =4$

$\sf \rightarrow f = - 2$

$\sf \: And \: also,$

$\sf c = - 36$

$\therefore \sf Radius \: of \: the \: circle \: (r) = \sqrt{ {g}^{2} + {f}^{2} - c}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ {3}^{2} + {2}^{2} - ( - 36 )}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{9 + 4 + 36}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{49}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 7$

$\sf \red{Therefore, \: the \: radius \: of \: the \: circle \: is \: 7 \: units \: and} \\ \sf \red{the \: center \: of \: the \: circle \: is \: (3, -2).}$