Answer
Verified
424.2k+ 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 $ .
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 $ .
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE