Search any question & find its solution
Question:
Answered & Verified by Expert
If the lines $3 x-4 y-7=0$ and $2 x-3 y-5=0$ are two diameters of a circle of area $49 \pi$ square units, the equation of the circle is
Options:
Solution:
2778 Upvotes
Verified Answer
The correct answer is:
$x^2+y^2-2 x+2 y-47=0$
$x^2+y^2-2 x+2 y-47=0$
Point of intersection of $3 x-4 y-7=0$ and $2 x-3 y-5=0$ is $(1,-1)$, which is the centre of the circle and radius $=7$.
$\therefore$ Equation is $(x-1)^2+(y+1)^2=49 \Rightarrow x^2+y^2-2 x+2 y-47=0$
$\therefore$ Equation is $(x-1)^2+(y+1)^2=49 \Rightarrow x^2+y^2-2 x+2 y-47=0$
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.