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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
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B
58
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C
87
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D
85
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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 circle
And $$d$$ 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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