ILMU PELAJARAN DAN PENDIDIKAN SEKOLAH: PERSAMAAN LINGKARAN dalam MATEMATIKA

Sabtu, 07 Agustus 2010

PERSAMAAN LINGKARAN dalam MATEMATIKA

Persamaan lingkaran yang mempunyai diameter ruas garis AB, dengan $A(x_{1},y_{1})$ dan $B(x_{2},y_{2})$ adalah $(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$

Bukti:

misal pusat lingkaran adalah P, maka titik P ditengah ruas garis AB, $P\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$ , karena ruas garis AB adalah diamter, maka $2r=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$ .sehingga $r^{2}=\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{4}$ .

Persamaan lingkaran dengan pusat P dan jari-jari r adalah:

$(x-\frac{x_{1}+x_{2}}{2})^{2}+(y-\frac{y_{1}+y_{2}}{2})^{2}=r^{2}$

$(x-\frac{x_{1}+x_{2}}{2})^{2}+(y-\frac{y_{1}+y_{2}}{2})^{2}=\frac{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}{4}$

$x^{2}-(x_{1}+x_{2})x+(\frac{x_{1}+x_{2}}{2})^{2}+y^{2}-(y_{1}+y_{2})y+(\frac {y_{1}+y_{2}}{2})^{2}= \frac{(x_{1}-x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}$

$x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(x_{1}-x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}-(\frac {x_{1}+x_{2}}{2})^{2}-(\frac {y_{1}+y_{2}}{2})^{2}$

$x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(x_{1}-x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}-\frac {(x_{1}+x_{2})^{2}}{4}-\frac {(y_{1}+y_{2})^{2}}{4}$

$x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y =\frac{(x_{1}-x_{2})^{2}}{4}-\frac {(x_{1}+x_{2})^{2}}{4}+\frac{(y_{1}-y_{2})^{2}}{4}-\frac {(y_{1}+y_{2})^{2}}{4}$

$x^{2}-(x_{1}+x_{2})x+y^{2}-(y_{1}+y_{2})y = \frac{(x_{1}-x_{2}+x_{1}+x_{2})(x_{1}-x_{2}-x_{1}-x_{2})}{4}+\frac{(y_{1}-y_{2}+y_{1}+y_{2})(y_{1}-y_{2}-y_{1}-y_{2})}{4}$