Find the center and radius of the circle given by the equation:

\[ x^2 + y^2 + 14x + 10y = 7 \]

( x − a ) 2 + ( y − b ) 2 = r 2 ( a ; b ) − t h e coor d ina t es o f t h e ce n t er r − t h e r a d i u s ================================== U se : ( k + l ) 2 = k 2 + 2 k l + l 2 ( ∗ ) ==================================
x 2 + y 2 + 14 x + 10 y = 7 x 2 + 14 x + y 2 + 10 y = 7 x 2 + 2 x ⋅ 7 + y 2 + 2 y ⋅ 5 = 7 U se ( ∗ ) x 2 + 2 x ⋅ 7 + 7 2 ​ ​ − 7 2 + U se ( ∗ ) y 2 + 2 y ⋅ 5 + 5 2 ​ ​ − 5 2 = 7 ( x + 7 ) 2 − 49 + ( y + 5 ) 2 − 25 = 7 ( x + 7 ) 2 + ( y + 5 ) 2 − 74 = 7 ∣ a dd 74 t o \both s i d es ( x + 7 ) 2 + ( y + 5 ) 2 = 81 ( x + 7 ) 2 + ( y + 5 ) 2 = 9 2
A n s w er : ( − 7 ; − 5 ) − ce n t er ; 9 − r a d i u s ​

Answered by Anonymous | 2024-06-10

( x − a ) 2 + ( y − b ) 2 = r 2
x 2 + y 2 + 14 x + 10 y = 7 x 2 + 14 x + 49 + y 2 + 10 y + 25 − 74 = 7 ( x + 7 ) 2 + ( y + 5 ) 2 = 81 a = − 7 , b = − 5 , r 2 = 81 ⇒ center = ( − 7 , − 5 ) , r = 81 ​ = 9

Answered by konrad509 | 2024-06-10

The center of the circle given by the equation is (-7, -5) and the radius is 9.
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Answered by Anonymous | 2024-11-06