Question

~QUIZ~
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Soal:
Berapakah titik pusat dan jari-jari dari persamaan lingkaran x² + y² - 4x + 2y + c = 0 yang melalui titik (5, -1)?
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Answer 1

Penjelasan dengan langkah-langkah:

x²+y²-4x+2y+c=0

5²+(-1)²-4(5)+2(-1)+c=0

25+1-20-2+c=0

c=-4

x²+y²+Ax+By+C=0

x²+y²-4x+2y-4=0

Pusat (-1/2A , -1/2B)

(-1/2(-4) , -1/2(2) )

(2,-1)

Jari jari √1/4A²+1/4B²-C

√1/4(-4)²+1/4(2)²-(-4)

√1/4(16)+1/4(4)+4

√4+1+4

√9

3

Answer 2

Verified answer

(5 => X , -1 => Y)

MAKA :

x² + y² - 4x + 2y + c = 0

(5²) + (-1)² - (4(5) + (2(-1) + c = 0

(5(5) + (-1(-1) - 20 + (-2) + c = 0

25 + (1(1) - 20 - 2 + c = 0

25 + 1 - 20 - 2 + c = 0

26 - 20 - 2 + c = 0

6 - 2 + c = 0

4 + c = 0

c = 0 - 4

c = -4

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x² + y² - 4x + 2y - 4 = 0

TP :

= (-1/2(A) , (-1/2(B)

= (-1/2(-4) , (-1/2(2)

= (-1(-4/2) , (-2/2)

= (-1(-2) , (-1)

= (1(2) , (-1)

= (2 , -1)

R = √(1/4(A²) + (1/4(B²) - C

= √(1/4(-4)²) + (1/4(2)²) - (-4)

= √(1/4(-4(-4) + (1/4(2(2) + 4

= √1/4(4(4) + (1/4(4) + 4

= √1/4(16) + (1(4/4) + 4

= √1(16/4) + 1 + 4

= √(1(4) + 1 + 4

= √4 + 1 + 4

= √5 + 4

= √9

= √(3(3)

= √3²

= 3