Question

find the radius of the circle 45x2+45y2-60+36y+19=0
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Answer 1

CORRECT QUESTION :–

• Find the radius of the circle 45x² + 45y² - 60x + 36y + 19 = 0.

ANSWER :–

GIVEN :–

• Equation of circle –

$\\ \implies\bf 45x^{2} + 45y^{2} - 60x + 36y + 19 = 0$

TO FIND :–

• Radius of circle = ?

SOLUTION :–

$\\ \implies\bf 45x^{2} + 45y^{2} - 60x + 36y + 19 = 0$

• We know that standard equation of circle is –

$\\ \implies\bf x^{2} + y^{2} + 2gx + 2fy +c= 0$

• And radius –

$\\ \large\implies\bf r = \sqrt{ {g}^{2} + {f}^{2} - c}$

• So that –

$\\ \implies\bf 45x^{2} + 45y^{2} - 60x + 36y + 19 = 0$

• We should write this as –

$\\ \implies\bf x^{2}+y^{2} - \dfrac{60}{45}x + \dfrac{36}{45}y + \dfrac{19}{45}= 0$

• Now compare –

$\\ \implies\bf g = - \dfrac{30}{45} = - \dfrac{2}{3}$

$\\ \implies\bf f= \dfrac{18}{45} = \dfrac{2}{5}$

$\\ \implies\bf c= \dfrac{19}{45}$

• Now put the values –

$\\ \large\implies\bf r = \sqrt{ { \bigg(- \dfrac{2}{3} \bigg)}^{2} + {\bigg(\dfrac{2}{5} \bigg)}^{2}-\dfrac{19}{45}}$

$\\ \large\implies\bf r = \sqrt{ { \bigg(\dfrac{4}{9} \bigg)}+ {\bigg(\dfrac{4}{25} \bigg)}-\dfrac{19}{45}}$

$\\ \large\implies\bf r = \sqrt{ { 4\bigg(\dfrac{1}{9} + \dfrac{1}{25} \bigg)}-\dfrac{19}{45}}$

$\\ \large\implies\bf r = \sqrt{ { 4\bigg( \dfrac{25 +9}{225} \bigg)}-\dfrac{19}{45}}$

$\\ \large\implies\bf r = \sqrt{ { 4\bigg( \dfrac{34}{225} \bigg)}-\dfrac{19}{45}}$

$\\ \large\implies\bf r = \sqrt{ { \dfrac{136}{225}}-\dfrac{19}{45}}$

$\\ \large\implies\bf r = \sqrt{ { \dfrac{6120 - 4275}{10125}}}$

$\\ \large\implies\bf r = \sqrt{ {\dfrac{1845}{10125}}}$

$\\ \large\implies\bf r = \sqrt{0.18}$

$\\ \large\implies{ \boxed{\bf r = 0.42}}$