Find the centre and radius of the circle -x - 5-displaystyle -2- - -y - 3-displaystyle -2- - 36

Question

Toppr Admin · Accepted Answer

The equation of the given circle is- -x-5-2-y-x2212-3-2-36-x21D2-x-x2212-x2212-5-2-y-x2212-3-2-62 which is of the form -x-x2212-h-2-y-x2212-k-2-r2 where h-x2212-5-k-3 and r-6Thus the centre of the given circle is -x2212-5-3- while its radius is 6

Find the centre and radius of the circle $(x + 5)^{2} + (y - 3)^{2} = 36$

The equation of the given circle is,
$(x + 5)^{2} + (y - 3)^{2} = 36$
$\Rightarrow {x - (- 5)}^{2} + {(y - 3)}^{2} = 6^{2}$
which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$
where $h = - 5, k = 3$ and $r = 6$
Thus the centre of the given circle is $(- 5, 3)$ while its radius is $6$

Find the centre and radius of the circle (x+5)2+(y−3)2=36

The equation of the given circle is, (x+5)2+(y−3)2=36⇒{x−(−5)}2+(y−3)2=62 which is of the form (x−h)2+(y−k)2=r2 where h=−5,k=3 and r=6Thus the centre of the given circle is (−5,3) while its radius is 6

Find the centre and radius of the circle $(x + 5)^{2} + (y - 3)^{2} = 36$

The equation of the given circle is,
$(x + 5)^{2} + (y - 3)^{2} = 36$
$\Rightarrow {x - (- 5)}^{2} + {(y - 3)}^{2} = 6^{2}$
which is of the form $(x - h)^{2} + (y - k)^{2} = r^{2}$
where $h = - 5, k = 3$ and $r = 6$
Thus the centre of the given circle is $(- 5, 3)$ while its radius is $6$