[Q.] If one end of the diameter of the circleis the point (–3, 2), t

1.	[Q.] If one end of the diameter of the circleis the point (–3, 2), then its other end is the point ?● Correct answer is:- ( 11 , 12) I want only steps....please give solutions.
Answer» Question :- ⇒ If one end of the DIAMETER of the circle $x^{2} + y^{2} -8x - 14y + c = 0$ is the point (–3, 2), then its other end is the point ? Answer:- ⇒ As , General equation of circle is ≡ $x^{2} + y^{2} + 2gx + 2fy + K = 0$ -----(1) ⇒ one end of circle is is B: ( -3 , 2 ) and other is A: ( a , b ).. ⇒ As , given equation of circle is ≡ $x^{2} + y^{2} -8x - 14y + c = 0$ comparing with equation (1).. We get , $g = - <klux>4</klux>$ and $f = - 7$ ⇒ As, we KNOW that CENTER of circle is : C = ( -g , -f ) ⇒ So, center is C : ( 4 , 7 ). So, by using mid point ⇒ $( 4 , 7 ) =( \frac{a-3}{2} , \frac{b + 2}{2} )$ ∴ 8 = a - 3 . ⇒ a = 11 ∴ 14 = b + 2 . ⇒ b = 12 So, another point is $A(a,b) = ( 11,12 )$ I hope this will help you thank you.

[Q.] If one end of the diameter of the circleis the point (–3, 2), then its other end is the point ?● Correct answer is:- ( 11 , 12) I want only steps....please give solutions.

Answer»

Question :-

⇒ If one end of the DIAMETER of the circle $x^{2} + y^{2} -8x - 14y + c = 0$ is the point (–3, 2), then its other end is the point ?

Answer:-

⇒ As , General equation of circle is

≡ $x^{2} + y^{2} + 2gx + 2fy + K = 0$ -----(1)

⇒ one end of circle is is B: ( -3 , 2 ) and other is A: ( a , b )..

⇒ As , given equation of circle is

≡ $x^{2} + y^{2} -8x - 14y + c = 0$

comparing with equation (1)..

We get ,

$g = - <klux>4</klux>$ and $f = - 7$

⇒ As, we KNOW that CENTER of circle is : C = ( -g , -f )

⇒ So, center is C : ( 4 , 7 ).

So, by using mid point

⇒ $( 4 , 7 ) =( \frac{a-3}{2} , \frac{b + 2}{2} )$

∴ 8 = a - 3 . ⇒ a = 11

∴ 14 = b + 2 . ⇒ b = 12

So,

another point is $A(a,b) = ( 11,12 )$