Question

Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4)

Answer 1

$\mathfrak{\bf{\underline{\underline{Gívєn \ :-}}}}$

ㅤㅤㅤㅤthє єquαtíσn σf thє círclє pαѕѕíng thrσugh

ㅤㅤㅤㅤㅤㅤ( 3,4 ) hαvíng cєntrє αt ( -3,4 ).

$\mathfrak{\bf{\underline{\underline{Tσ \ fínd \ :-}}}}$

ㅤㅤㅤㅤㅤㅤㅤThє єquαtíσn σf círclє fσr gívєn dαtα.

$\mathfrak{\bf{\underline{\underline{Sσlutíσn \ :-}}}}$

wє knσw thαt,

єquαtíσn σf círclє :—

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

whєrє,

$(x_{1},y_{1}) \: are \: the \: co - ordinates \: of \: centre \: of \\ \: the \: circle \: and \: r \: is \: the \: radius$

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

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

$\sqrt{[3 - ( - 3)] ^{2} +[ (4 - 4)] ^{2} } = r$

$ㅤㅤㅤㅤ \sqrt{( {3})^{2} } = r$

$ㅤㅤㅤㅤㅤㅤㅤ \boxed{r = 3}$

hєncє thє єquαtíσn σf thє círclє íѕ,

$\implies \: [(x - ( - 3) ]^{2} + [(y - 4)] ^{2} = {3}^{2}$

$\implies \: (x + 3) ^{2} + (y - 4) ^{2} = 9$

$\implies \: ({x}^{2} + 6x + 9) + ( {y}^{2} - 8y + 16) = 9$

$\implies \: \: \: \: \: \: {x}^{2} + {y}^{2} + 6x - 8x + 16 = 0$

$\implies \: \: \: \: \: \boxed{ \boxed{ {x}^{2} + {y}^{2} + 6x - 8y + 16 = 0}}$