Question

Persamaan lingkaran x2 + y2 -mx + 7y + 4 = 0 menyinggung sumbu x maka nilai m adalah

Answer 1

Verified answer

Lingkaran

x² + y² - mx + 7y + 4 = 0
(x - m/2)² + (y + 7/2)² = -4 + (m/2)² + (7/2)²
P(m/2 , -7/2)

Menyinggung sb x ---> y = 0
r = 7/2

r² = -4 + m²/4 + 49/4
(7/2)² = -4 + m²/4 + 49/4
0 = -4 + m²/4
m²/4 = 4
m² = 16
m = 4 atau m = -4

Answer 2

Verified answer

Jawab

x² + y² - mx+ 7y + 4 = 0 --> x² +y² + Ax +By +C =0

P= (-1/2 A, - 1/2 B)  → Pusat  (1/2m, - 7/2)
r = √(-1/2 A)²+(-1/2B)²- C) → r1 = √(1/2 m)²+ (7/2)² - 4) 

menyinggung sumbu x , y = 0 -->  r 1= 7/2

r1= r1
√(1/2m)² + (7/2)²- 4) = 7/2
1/4 m² + 49/4 - 4 = 49/4
1/4 m² = 4
m²= 16
m = 4 atau m = - 4