[Q.] If one end of the diameter of the circle
[tex] {x}^{2} + {y}^{2} – 8x – 14y + c = 0[/tex]
is the point (–3, 2),

1 thought on “[Q.] If one end of the diameter of the circle <br />[tex] {x}^{2} + {y}^{2} – 8x – 14y + c = 0[/tex]<br />is the point (–3, 2),”

1. Question :-

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

   Answer:-

   ⇒ As , General equation of circle is

   ≡ [tex]x^{2} + y^{2} + 2gx + 2fy + K = 0[/tex] —–(1)

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

   ⇒ As , given equation of circle is

   ≡ [tex]x^{2} + y^{2} -8x – 14y + c = 0[/tex]

   comparing with equation (1)..

   We get ,

   [tex]g = – 4[/tex] and [tex]f = – 7[/tex]

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

   ⇒ So, center is C : ( 4 , 7 ).

   So, by using mid point

   ⇒ [tex]( 4 , 7 ) =( \frac{a-3}{2} , \frac{b + 2}{2} )[/tex]

   ∴ 8 = a – 3 . ⇒ a = 11

   ∴ 14 = b + 2 . ⇒ b = 12

   So,

   another point is [tex]A(a,b) = ( 11,12 )[/tex]

   • I hope this will help you
   • thank you.

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