Question

The radius of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?

Answer 1

$\textbf{Given:}$

$\textsf{Equation of circle is}$

$\mathsf{4x^2+4y^2-8x+12y-25=0}$

$\textbf{To find:}$

$\textsf{Radius of the circle}$

$\textbf{Solution:}$

$\underline{\textsf{Concept:}}$

$\mathsf{Radius\;of\;the\;circle\;x^2+y^2+2gx+2fy+c=0\;is}$

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

$\mathsf{Consider}$

$\mathsf{4x^2+4y^2-8x+12y-25=0}$

$\textsf{Dividing bothsides by 4}$

$\mathsf{x^2+y^2-2x+3y-\dfrac{25}{4}=0}$

$\mathsf{Here,\;2g=-2,\;2f=3,\;c=\dfrac{-25}{4}}$

$\implies\mathsf{g=-1,\;f=\dfrac{3}{2},\;c=\dfrac{-25}{4}}$

$\mathsf{Radius=\sqrt{g^2+f^2-c}}$

$\mathsf{=\sqrt{(-1)^2+(\frac{3}{2})^2+\dfrac{25}{4}}}$

$\mathsf{=\sqrt{1+\dfrac{9}{4}+\dfrac{25}{4}}}$

$\mathsf{=\sqrt{\dfrac{38}{4}}}$

$\mathsf{=\dfrac{\sqrt{38}}{2}}$

$\textbf{Answer:}$

$\mathsf{Radius\;of\;the\;circle\;is\;\dfrac{\sqrt{38}}{2}\;units}$

$\textbf{Find more:}$

If the circle

x² + y2 - 4x +6y+a = 0 has

radius 4 then find a ,b

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Answer 2

The radius of circle 4x² + 4y² - 8x + 12y - 25 = 0 is $\displaystyle \bf{ \frac{ \sqrt{38} }{2} \: \: unit }$

Given :

The equation of the circle 4x² + 4y² - 8x + 12y - 25 = 0

To find :

The radius of the circle

Solution :

Step 1 of 2 :

Write down the given equation of the circle

The equation of the circle is

4x² + 4y² - 8x + 12y - 25 = 0

Step 2 of 2 :

Find radius of the circle

$\displaystyle \sf{ 4 {x}^{2} + 4 {y}^{2} - 8x + 12y - 25 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + {y}^{2} - 2x + 3y - \frac{25}{4} = 0 }$

Comparing with the general equation of the circle

x² + y² + 2gx + 2fy + c = 0

$\displaystyle \sf{ g = - 1 \: , \: f = \frac{3}{2} \: , \: c = - \frac{25}{4} }$

Hence radius of the circle

$\displaystyle \sf{ = \sqrt{ {g}^{2} + {f}^{2} - c} }$

$\displaystyle \sf{ = \sqrt{ {( - 1)}^{2} + { \bigg( \frac{3}{2} \bigg)}^{2} - \bigg( - \frac{25}{4} \bigg)} }$

$\displaystyle \sf{ = \sqrt{1 + \frac{9}{4} + \frac{25}{4} } }$

$\displaystyle \sf{ = \sqrt{ \frac{4 + 9 + 25}{4} } }$

$\displaystyle \sf{ = \sqrt{ \frac{38}{4} } }$

$\displaystyle \sf{ = \frac{ \sqrt{38} }{2} }$

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