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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Circle

BITSAT 2011 · Mathematics · Circle

For what value of k the circles x²+y²+5 x+3 y+7=0 and x²+y²-8 x+6...

For what value of kk the circles x2+y2+5x+3y+7=0x²+y²+5 x+3 y+7=0 and x2+y2−8x+6y+k=0x²+y²-8 x+6 y+k=0 cuts orthogonally

Options

  1. A16
  2. B-18
  3. C-13
  4. D-10

Correct answer

B. -18

Step-by-step solution

Let the two circles be x2+y2+2g1x+2f1yx²+y²+2 g_1 x+2 f_1 y +c1=0and+c_1=0 andx²+y²+2 g_2 x+2 f_2 y+c_2=0wherewhereg_1=5 / 2, f_1=3 / 2, c_1=7, g_2=4, f_2=3and(c2=and (\mathrm{c}_{2}=\mathrm{k}Ifthetwocirclesintersectsorthogonally,thenIf the two circles intersects orthogonally, then2(g1g2+f1f2)=c1+c2⇒2(−10+92)=7+k⇒11=7+k⇒k=−18\begin{aligned} & 2 \left(\mathrm{g}_{1} \mathrm{ g}_{2}+ \mathrm{f}_{1} \mathrm{f}_{2}\right)= \mathrm{c}_{1}+\mathrm{c}_{2} \Rightarrow 2\left(-10+\frac{9}{2}\right)=7+\mathrm{k}\\ & \Rightarrow 11=7+\mathrm{k} \Rightarrow \mathrm{k}=-18 \end{aligned}$)
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