Question

[Q.] 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 ?

● Correct answer is:- ( 11 , 12)
I want only steps....
please give solutions. ​

Answer 1

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 ,

           $g = - 4$ 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.