Search any question & find its solution
Question:
Answered & Verified by Expert
The pole of the straight line $9 x+y-28=0$ with respect to circle $2 x^2+2 y^2-3 x+5 y-7=0$, is
Options:
Solution:
1306 Upvotes
Verified Answer
The correct answer is:
$(3,-1)$
$\begin{aligned} & x^2+y^2-\frac{3}{2} x+\frac{5}{2} y-\frac{7}{2}=0 \\ & \Rightarrow\left(x-\frac{3}{4}\right)^2+\left(y+\frac{5}{4}\right)^2-\frac{9}{16}-\frac{25}{16}-\frac{7}{2}=0 \\ & \Rightarrow\left(x-\frac{3}{4}\right)^2+\left(y+\frac{5}{4}\right)^2-\frac{45}{8}=0\end{aligned}$
Put $X=x-\frac{3}{4}$ and $Y=y+\frac{5}{4}$, we get the equation of circle $X^2+Y^2-\frac{45}{8}=0$ and the line $9 X+Y-\frac{45}{2}=0$
Hence pole$\equiv\left(\frac{9 \times \frac{45}{8}}{\frac{45}{2}}, \frac{1 \times \frac{45}{8}}{\frac{45}{2}}\right) \equiv\left(\frac{9}{4}, \frac{1}{4}\right)$
$x=\frac{9}{4}+\frac{3}{4} \text { and } y=\frac{1}{4}-\frac{5}{4}=-1$
Hence the pole is $(3,-1)$.
Put $X=x-\frac{3}{4}$ and $Y=y+\frac{5}{4}$, we get the equation of circle $X^2+Y^2-\frac{45}{8}=0$ and the line $9 X+Y-\frac{45}{2}=0$
Hence pole$\equiv\left(\frac{9 \times \frac{45}{8}}{\frac{45}{2}}, \frac{1 \times \frac{45}{8}}{\frac{45}{2}}\right) \equiv\left(\frac{9}{4}, \frac{1}{4}\right)$
$x=\frac{9}{4}+\frac{3}{4} \text { and } y=\frac{1}{4}-\frac{5}{4}=-1$
Hence the pole is $(3,-1)$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.