Question

If the circle
x² + y2 - 4x +6y+a = 0 has
radius 4 then find a ,b

Answer 1

$\textbf{Given:}$

$\text{Equation of circle is x^2+y^2-4x+6y+a=0 }$

$\textbf{To find:}$

$\text{The values of 'a'}$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{The radius of the circle x^2+y^2+2gx+2fy+c=0 is}$

$\bf\sqrt{g^2+f^2-c}$

$\text{Comparing x^2+y^2-4x+6y+a=0 with}$

$\text{ x^2+y^2+2gx+2fy+c=0 we get g=-2, f=3 and c=a}$

$\textbf{Radius = 4 units}$

$\implies\sqrt{g^2+f^2-c}=4$

$\implies\,g^2+f^2-c=16$

$\implies\,(-2)^2+3^2-a=16$

$\implies\,4+9-a=16$

$\implies\,-a=16-13$

$\implies\bf\,a=-3$

$\therefore\textbf{The value of a is -3}$

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Answer 2

Step-by-step explanation:

\text{We know that,}We know that,

\text{The radius of the circle $x^2+y^2+2gx+2fy+c=0$ is}The radius of the circle x

2

+y

2

+2gx+2fy+c=0 is

\bf\sqrt{g^2+f^2-c}

g

2

+f

2

−c

\text{Comparing $x^2+y^2-4x+6y+a=0$ with}Comparing x

2

+y

2

−4x+6y+a=0 with

\text{$x^2+y^2+2gx+2fy+c=0$ we get g=-2, f=3 and c=a}x

2

+y

2

+2gx+2fy+c=0 we get g=-2, f=3 and c=a

\textbf{Radius = 4 units}Radius = 4 units

\implies\sqrt{g^2+f^2-c}=4⟹

g

2

+f

2

−c

=4

\implies\,g^2+f^2-c=16⟹g

2

+f

2

−c=16

\implies\,(-2)^2+3^2-a=16⟹(−2)

2

+3

2

−a=16

\implies\,4+9-a=16⟹4+9−a=16

\implies\,-a=16-13⟹−a=16−13

\implies\bf\,a=-3⟹a=−3

\therefore\textbf{The value of a is -3}∴The value of a is -