Question

Which of the following lines is a diameter of circle ${{x}^{2}}+y{}^{2}-6x-8y-9=0$?
A . .$3x-4y=0$
B. $4x-y=0$
C. $x+y=7$
D. $x-y=1$

Answer 1

Hint: In this question, we are given an equation of circle and we have to find the diameter of the circle. To solve this, first we compare the given equation with the equation of circle and there we get the value of g,f and c. We know diameter passes through the centre. So from the given options, we put the centre of the circle and find out the option which satisfies the centre of the circle.

Formula Used:
General form of circle :- ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

Complete Step – By - Step Solution:
We have given the equation of circle ${{x}^{2}}+y{}^{2}-6x-8y-9=0$--------------- (1)
We have to find the diameter of circle.
We know general form of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$---------------- (2)
Now we compare the equation (1) with equation (2), we get
$2g=-6$ and $2f=-8$ and $c=-9$
That is $g=-3$, $f=-4$ and $c=-9$
Now we know centre of circle is (-g, -f) i.e. ( 3,4)
As the diameter passes of the circle will pass through its centre.
From the given options,
 Only $x+y=7$ is passing through the centre ( 3,4 )

Thus, Option ( C ) is correct.

Note: Another method to solve this question is as follow:
Alternate Method:
 Given equation of circle ${{x}^{2}}+y{}^{2}-6x-8y-9=0$
$({{x}^{2}}-6x+9)+({{y}^{2}}-8y+16)-9-9-16=0$
${{(x-3)}^{2}}+(y-4){}^{2}-34=0$
That is ${{(x-3)}^{2}}+(y-4){}^{2}=34$
On comparing the equation with${{(x-h)}^{2}}+(y-k){}^{2}={{r}^{2}}$, we get
$h=3,k=4$
As the diameter passes of the circle will pass through its centre.
From the given options,
 Only $x+y=7$ is passing through the centre ( 3,4 )