The radical axis of the circles 3x^2 + 3y^2 - | vedclass

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(B) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,rewrite the equations in the form $x^2 + y^2 + 2gx + 2fy

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$x + 10y - 2 = 0$

Vedclass · Answer

(B) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,rewrite the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.The first circle is $x^2 + y^2 - \frac{7}{3}x + \frac{8}{3}y + \frac{11}{3} = 0$.The second circle is $x^2 + y^2 - 3x - 4y + 5 = 0$.The radical axis is $\left( -\frac{7}{3} - (-3) \right)x + \left( \frac{8}{3} - (-4) \right)y + \left( \frac{11}{3} - 5 \right) = 0$.This simplifies to $\left( -\frac{7}{3} + \frac{9}{3} \right)x + \left( \frac{8}{3} + \frac{12}{3} \right)y + \left( \frac{11-15}{3} \right) = 0$.$\frac{2}{3}x + \frac{20}{3}y - \frac{4}{3} = 0$.Multiplying by $\frac{3}{2}$,we get $x + 10y - 2 = 0$.

The radical axis of the circles $3x^2 + 3y^2 - 7x + 8y + 11 = 0$ and $x^2 + y^2 - 3x - 4y + 5 = 0$ is

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