Lingkaran yang berpusat di (-2, 4) dan menyingung garis 3x - 4y -3 =0 adalah??.... Question from @Cheveux

feedback54 $(x+2)^2+(y-4)^2=r^2 \\ x^2+4x+4+y^2-8y+16=r^2 \\ y= \frac{3}{4} (x-1) \\ \\ x^2+4x+4+(\frac{3}{4} (x-1))^2-8(\frac{3}{4} (x-1))+16=r^2 \\ x^2+4x+4+\frac{9}{16} (x^2-2x+1)-6 (x-1)+16=r^2 \\ x^2+4x+4+\frac{9}{16}x^2-\frac{9}{8}x+\frac{9}{16}-6x+6+16=r^2 \\ (\frac{9}{16}+1)x^2-\frac{25}{8}x+26+\frac{9}{16}-r^2=0 \\ \\ D=0 \\ \frac{625}{64}-4(\frac{9}{16}+1)(26+\frac{9}{16}-r^2)=0 \\ \frac{625}{64}-4(\frac{25}{16})(\frac{425-16r^2}{16})=0 \\ \frac{625}{64}-(\frac{25}{4})(\frac{425-16r^2}{16})=0$
$25-425+16r^2=0 \\ 16r^2=400 \\ r^2=25$

$x^2+4x+4+y^2-8y+16=r^2 \\ x^2+4x+4+y^2-8y+16=25 \\ x^2+y^2+4x-8y-5=0$

0 votes Thanks 3

feedback54 Ada yang tau cara yang lebih simple gak?

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