Question

Kuis:
.....................

Answer 1

persamaan garis singgung lingkaran

Titik singgung P(-1,4)

r = √((-1)² + 4²) = √17

gradien garis normal, m1 = y/x = 4/(-1) = -4

gradien garis singgung, m2 = -1/m1 = 1/4

PGSL dg m = 1/4 :

y = mx ± r√(1 + m²)

y = 0,25x ± √17 × √(1 + (1/4)²)

y = 0,25x ± √17 × 1/4 √17

y = 0,25x ± 17/4

y = 0,25x ± 4,25

PGSL di P(-1,4) :

y = 0,25x + 4,25

Answer 2

Penjelasan dengan langkah-langkah:

given :

• P (-1,4) [tex](x_1=-1\:,\:y_1=4)[/tex]
• circle equation: x²+y²=r² (because the center of the circle is (0,0))

asked: the equation of the tangent to the circle at P

answer :

formula to find equation of the tangent to the circle is :

[tex]\rm\boxed{x.x_1+y.y_1=r^2}[/tex]

then we should find a r (radius)

we can find a r (radius) use pythagoras formula

r = [tex]\sqrt{x_1^2+y_1^2}[/tex]

r = [tex]\sqrt{(-1)^2+(4)^2}[/tex]

r = [tex]\sqrt{1+16}[/tex]

r = [tex]\sqrt{17}[/tex]

So the equation of the tangent to the circle is

[tex]x. x_1 + y.y_1 = {r}^{2} [/tex]

[tex]x. - 1 + y.4 = ( \sqrt{17})^{2} [/tex]

[tex] - 1x + 4y = 17[/tex]

change that equation in the y=mx+c

-1x + 4y = 17

4y = 1x + 17

y = ¼x + ¼17

•Then the equation of the tangent to the circle at P is y = ¼x + ¼17