Question

If the circles $$2{x}^{2}+2{y}^{2}+{p}{x}+6{y}-10=0$$ and $$3{x}^{2}+3{y}^{2}+15{x}+ py +21=0$$ are orthogonal, then the value of $$p$$ is

A
78
B
58
C
87
D
85

Solution

The correct option is B $$\dfrac87$$If two circles are orthogonal , then$$r_{1}^{2}+r_{2}^{2}=d^{2}$$where, $$r_{1}$$ and $$r_{2}$$ are radias of circleAnd $$d$$ is distance between circles.$$\therefore \displaystyle \left[\left(\frac{p}{4}\right)^{2}+\left(\frac{3}{2}\right)^{2}+5\right]+\left[\left(\frac{5}{2}\right)^{2}+\left(\frac{p}{6}\right)^{2}-7\right]=\left(\frac{p}{4}-\frac{5}{2}\right)^{2}+\left(\frac{3}{2}-\frac{p}{6}\right)^{2}$$$$-2=-\dfrac{5p}{4}-\dfrac{P}{2}$$$$\Rightarrow -2=\dfrac{-7p}{4}$$$$\Rightarrow P=\dfrac{8}{7}$$Hence, option C.Maths

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