Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

In the equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ represents a circle with X-axis as a diameter and radius a, then which of the following is possible.
A. $ f=2a,g=0,c=3{{a}^{2}} $
B. $ f=0,g=a,c=3{{a}^{2}} $
C. $ f=0,g=-2a,c=3{{a}^{2}} $
D. none of these

seo-qna
Last updated date: 19th Jul 2024
Total views: 395.1k
Views today: 9.95k
Answer
VerifiedVerified
395.1k+ views
Hint: We first try to form the given circle in its general form of $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ to find the centre and the radius. from there we place the values of the y coordinate of the centre as 0. Then we put the value of $ y=0 $ in the equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ to get end points of the diameter. We put the possible given options and check the most appropriate one.

Complete step by step answer:
The equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ represents a circle with X-axis as a diameter and radius a.
As the diameter is a part of X-axis, it means that the centre is on the X-axis and the y coordinate of the centre is 0.
It’s given that the equation of the circle is $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ . We transform it in its general form of $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ and get $ {{\left( x-g \right)}^{2}}+{{\left( y-f \right)}^{2}}={{\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)}^{2}} $ . O is the centre.
Equating with the general equation of circle $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ , we get the centre as $ O\equiv \left( -g,-f \right) $ and the radius as $ \sqrt{{{g}^{2}}+{{f}^{2}}-c} $ units.
We can put the value $ -f=0 $ which means $ f=0 $ . Also, the radius is a.
This means the centre is $ O\equiv \left( -g,0 \right) $ and the radius as $ \sqrt{{{g}^{2}}+{{f}^{2}}-c}=a $ units.
Solving the radius, we get $ {{g}^{2}}-c={{a}^{2}} $ .
Now we put the value of $ y=0 $ in the equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ to get the two end points of the diameter. This gives $ {{x}^{2}}+2gx+c=0 $ . Solving we get
 $ x=\dfrac{-2g\pm \sqrt{4{{g}^{2}}-4c}}{2}=-g\pm \sqrt{{{g}^{2}}-c} $ .
Now we place the given two options which are possible and they are $ f=0,g=a,c=3{{a}^{2}} $ and $ f=0,g=-2a,c=3{{a}^{2}} $ . In both cases $ f=0,c=3{{a}^{2}} $ . Now we put $ c=3{{a}^{2}} $ in $ {{g}^{2}}-c={{a}^{2}} $ .
We get $ {{g}^{2}}=4{{a}^{2}} $ which gives $ g=\pm 2a $ .
Therefore, the correct option is C.

Note:
We need to remember that the general equation $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ always makes an intercept $ 2\sqrt{{{g}^{2}}-c} $ units on the X-axis and $ 2\sqrt{{{f}^{2}}-c} $ units on the Y-axis. We use that to find the diameter length in the circle of $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ .