Question

if x²+y²_4x+6y+c=0 represents a circle with radius 6 then find the value of c​

Answer 1

Answer:

x^2+y^2-4x+6y+c=0............(1)

the general equatiom pf circle is

(x+a)^2+(y+b)^2=r^2

where r is the radius of the circle

if we represent the eq 1 in this format we get

(x-2)^2+(y+3)^2-4-9+c=0

(x-2)^2+(y+3)^2=13-c

here 13-c=r^2

and given that r =6

hence 13-c=6×6

c=13-36

c=-23

Answer 2

Answer:

The value of c = - 23

Step-by-step explanation:

Given data

the equation x² + y² - 4x + 6y + c = 0_(1)  represents a circle

and radius of the circle r = 6 units

here we need to find the value of c  

compare the given equation (1) with general form of circle

                                                      x² + y² + 2gx + 2fy + c = 0

⇒  2g = - 4                      ⇒ 2f = 6          

       g = - 2                            f = 3  

here the radius of the circle R = $\sqrt{g^{2} + f^{2} - c }$  

                                                  = $\sqrt{(-2)^{2} + 3^{2} - c}$

                                                  = $\sqrt{4 + 9 - c }$

                                                  = $\sqrt{13 - c }$

from given data radius of circle = 6 units

⇒  $\sqrt{13 - c } = 6$

⇒    $(\sqrt{13-c})^{2}$  = 6²    [ squaring on b sides ]

⇒    13 - c = 36

⇒   c =  - 36 + 13

⇒   c = - 23