Question Number 127047 by mohammad17 last updated on 26/Dec/20
![how can graph this [x^2 +(y−1)^2 >9 ] pleas sir help me with details ?](Q127047.png)
$${how}\:{can}\:{graph}\:{this}\: \\ $$ $$\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\:\right]\:{pleas}\:{sir}\:{help}\:{me}\:{with}\:{details}\:? \\ $$
Answered by benjo_mathlover last updated on 26/Dec/20
$${if}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{3}\:,\:{the}\:{locus}\:{is}\:{a}\:{circle} \\ $$ $${with}\:{centre}\:{point}\:{at}\:\left(\mathrm{0},\mathrm{1}\right)\:{and}\:{radius}\:\sqrt{\mathrm{3}} \\ $$ $$ \\ $$
Commented bymohammad17 last updated on 26/Dec/20
$${sory}\:{sir}\:\left({x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\right) \\ $$
Commented bybenjo_mathlover last updated on 26/Dec/20
$${what}\:{the}\:{meaning}\:\left(\:\:\:\right)\:? \\ $$
Commented bymohammad17 last updated on 26/Dec/20
![[x^2 +(y−1)^2 >9]](Q127054.png)
$$\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\right] \\ $$
Commented bymohammad17 last updated on 26/Dec/20
Commented bymohammad17 last updated on 26/Dec/20
![sir how the graph [x^2 +(y−1)^2 >9]it is became alssoe](Q127058.png)
$${sir}\:{how}\:{the}\:{graph}\:\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\right]{it}\:{is}\:{became}\:{alssoe} \\ $$
Answered by ebi last updated on 26/Dec/20
$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9} \\ $$ $$\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}−{x}^{\mathrm{2}} \\ $$ $${y}−\mathrm{1}>\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\:\:{or}\:\:{y}−\mathrm{1}<−\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$ $${y}>\mathrm{1}+\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }\:\:{or}\:\:\:\:\:\:\:\:\:\:{y}<\mathrm{1}−\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$
Commented byebi last updated on 26/Dec/20
