Question Number 127047 by mohammad17 last updated on 26/Dec/20 | ||
$${how}\:{can}\:{graph}\:{this}\: \\ $$ $$\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\:\right]\:{pleas}\:{sir}\:{help}\:{me}\:{with}\:{details}\:? \\ $$ | ||
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 | ||
$$\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\right] \\ $$ | ||
Commented bymohammad17 last updated on 26/Dec/20 | ||
Commented bymohammad17 last updated on 26/Dec/20 | ||
$${sir}\:{how}\:{the}\:{graph}\:\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\right]{it}\:{is}\:{became}\:{alssoe} \\ $$ | ||
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 | ||