If x^2 + y^2 + px + 3y - 5 = 0 and x^2 + y^2 | vedclass

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(A) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f

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$\frac{1}{2}$

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(A) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Comparing the given equations with the standard form:For the first circle: $2g_1 = p \Rightarrow g_1 = \frac{p}{2}$,$2f_1 = 3 \Rightarrow f_1 = \frac{3}{2}$,$c_1 = -5$.For the second circle: $2g_2 = 5 \Rightarrow g_2 = \frac{5}{2}$,$2f_2 = p \Rightarrow f_2 = \frac{p}{2}$,$c_2 = 7$.Substituting these values into the condition:$2\left(\frac{p}{2}\right)\left(\frac{5}{2}\right) + 2\left(\frac{3}{2}\right)\left(\frac{p}{2}\right) = -5 + 7$$\frac{5p}{2} + \frac{3p}{2} = 2$$\frac{8p}{2} = 2$$4p = 2$$p = \frac{2}{4} = \frac{1}{2}$.

If $x^2 + y^2 + px + 3y - 5 = 0$ and $x^2 + y^2 + 5x + py + 7 = 0$ cut orthogonally,then $p$ is

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