diketahui persamaan lingkaran x²+y²+4x+6y-12=0. jika lingkaran tersebut di cerminkan terhadap garis x-y=0 dilanjutkan dilatasi D(0,-3),tentukan a.persamaan baya

jawab

x²+y²+ 4x + 6y -12= 0
(x+2)² +(y+3)² = 12 + 2²+3²
(x+2)² + (y+3)²= 25
P(-2,-3) , r= 5

T1 pencerminan garis x-y = 0 atau garis y = x

[tex]T _{1} \,= \left[\begin{array}{ccc}0&1\\1&0\end{array}\right] [/tex]


T2 Dilatasi [O, -3] → P(0,0), k = - 3

[tex]T _{2} \,= \left[\begin{array}{ccc}-3&0\\0&-3\end{array}\right][/tex]

a)  (x', y') = T2 o T1 (x,y) 

[tex] \left[\begin{array}{ccc}x'\\y'\end{array}\right] \,= \left[\begin{array}{ccc}-3&0\\0&-3\end{array}\right] \, \left[\begin{array}{ccc}0&1\\1&0\end{array}\right]\, \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] \,= \left[\begin{array}{ccc}0&-3\\-3&0\end{array}\right] \, \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] \,= \left[\begin{array}{ccc}-3y\\-3x\end{array}\right][/tex]

x' = - 3y → y = - 1/3 x'
y' = - 3x → x = - 1/3 y'
substitusi ke x²+y²+ 4x + 6y -12= 0
(-1/3 y)² + (-1/3 x)²+ 4( - 1/3 y) + 6(-1/3 x) - 12 = 0
1/9 y² + 1/9 x² - 4/3 y - 2x - 12=0..kali 9
bayangannya  adalah
x² + y² - 18 x - 12y - 108 = 0

b.   Perbandingan luas asal dan bayangan = 1 : k²
L1 : L2 = 1 : (-3)² = 1 : 9