Find the equation of circle whose diameters are 2x-3y+12=0 and x+4y-5=0 and area is 154sq units. (Use π=22/7).
Answers
Answer:
hey mate there is your answer
Step-by-step explanation:
As 2x - 3y =5 and 3x-4y=7 are the diameters of a circle
=> Their point of intersection is the center of circle
=> On solving equations we get,
2x-3y=5 ----(i) x 3
3x-4y=7 ----(ii) x 2
6x - 9y = 15 ---(iii)
6x - 8y = 14 ---(iv)
On subtracting equations (iv) from (iii) gives ,
( 6x - 9y ) - ( 6x - 8y ) = 15 - 14
=>6x - 9y - 6x + 8y = 1
=> - y = 1 => y = - 1,On putting the value of y in (i), gives
2x - 3(- 1) = 5
=> 2x + 3 = 5 => 2x = 5 - 3 => 2x = 2 => x = 1
Thus Center of a circle is ( x , y ) = ( 1 , -1 )
Now Area of circle = 154 sq. units
=> ( Pie ) ( r ^ 2 ) = 154
=> ( 22/7 ) x ( r ^2 ) = 154
=> r^2 = 154 x 7 / 22
= 14 x 7 / 2 ( on dividing 154 and 22 by 11)
= 7 x 7
=> r = 7 units
Finally, The equation of a circle having center ( 1 , - 1 ) and with radius r = 7 units is
\begin{lgathered}{(x - 1)}^{2} + {(y - ( - 1))}^{2} = {7}^{2} \\ {(x - 1)}^{2} + {(y + 1))}^{2} = 49 \\ {x}^{2} - 2x + 1 + {y}^{2} + 2y + 1 = 49 \\ {x}^{2} + {y}^{2} - 2x + 2y + 2 = 49 \\ {x}^{2} + {y}^{2} - 2x + 2y + 2 - 49 = 0 \\ {x}^{2} + {y}^{2} - 2x + 2y - 47 = 0\end{lgathered}
(x−1)
2
+(y−(−1))
2
=7
2
(x−1)
2
+(y+1))
2
=49
x
2
−2x+1+y
2
+2y+1=49
x
2
+y
2
−2x+2y+2=49
x
2
+y
2
−2x+2y+2−49=0
x
2
+y
2
−2x+2y−47=0