Question

Persamaan garis singgung pada grafik x2+y2+4x-2y-27=0 melalui titik D(2,-3) adalah​

Answer 1

x^2 + y^2 + 4x - 2y - 27 = 0

t.pusat = (-1/2 . a , -1/2 . b) = (-2,1)

r^2 = (-2)^2+(1)^2-(-27)

r^2 = 4 + 1 + 27

r^2 = 32

r = 4 akar 2

pgs melalui (2,-3) dengan pusat (-2,1)

(x+2)^2 +(y-1)^2 = (4 akar 2)^2

(2+2)(x+2) + ((-3)-1)(y-1) = 32

4(x+2) + (-4)(y-1) =32

(4x + 8 ) + (-4y + 4 ) = 32

4x - 4y + 12 = 32

4x - 4y = 20

x - y = 5

Answer 2

Jawaban

x - y - 5 = 0

Penjelasan

A = 4x

B = -2y

C = -27

titik D (2 , -3)

x1=2 , y1=-3

mencari titik pusat :

》( -1/2A , -1/2B)

》( -1/2×4 , -1/2×(-2))

》(-2 , 1)

mencari jari-jari :

r^2 = (-1/2A^2) + (-1/2B^2) - C

r^2 = (-2)^2 + (1)^2 - (-27)

r^2 = 4+1+27

r^2 = 32

rumus persamaan garis singgung lingkaran yg melalui pusat (a , b) dan melalui titik (x1 , y1) :

(x - a)(x1 - a) + (y - b)(y1 - b) = r^2

maka :

(x - a)(x1 - a) + (y - b)(y1 - b) = r^2

(x - (-2))(2 - (-2)) + (y - 1)(-3 - 1) = 32

(x + 2)(4) + (y - 1)(-4) = 32

4x + 8 - 4y + 4 = 32

4x - 4y + 12 - 32 = 0

4x - 4y - 20 = 0

x - y - 5 = 0

jadi persamaan garis singgungnya adalah x -

y - 5 = 0

smg membantu^^