Tolong bntu jawab
1.berapa radius dari persamaan lingkaran berikut? X²+Y²-10X+8Y-23=0
2.Dimanakah pusat dari lingkaran berikut?X²+Y²+4X+6Y-36=0
3.Dimanakah pusat lingkaran dengan persamaan (X+2)²+(Y-4)²=41
Answers
Given : Persamaan lingkaran circle equations i) X² + Y²-10X + 8Y-23 = 0 ii) X² + Y² + 4X + 6Y-36 = 0 iii) (X + 2) ² + (Y-4) ² = 41
To find : radius and center jari-jari dan pusat lingkaran
Solution :
1. What is the radius of the following circle equation? X² + Y²-10X + 8Y-23 = 0
2. Where is the center of the following circle? X² + Y² + 4X + 6Y-36 = 0
3. Where is the center of the circle with the equation (X + 2) ² + (Y-4) ² = 41
X² + Y²-10X + 8Y-23 = 0
= (x - 5)² - 25 + (y + 4)² -16 - 23 = 0
=> (x - 5)² + (y + 4)² = 64
=> (x - 5)² + (y + 4)² = 8²
(x - h)² + (y - k)² = r²
jari-jari dan pusat lingkaran = ( 5 , - 4) & 8
X² + Y² + 4X + 6Y-36 = 0
=> (x + 2)² - 4 + (y + 3)² - 9 - 36 = 0
=> (x + 2)² + (y + 3)² = 7²
( -2 , - 3) & 7
(X + 2) ² + (Y-4) ² = 41
(-2 , 4) & √41
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