write-the-equation-of-the-circle-which-containing-the-points-4-3-1-2-and-center-on-line-3x-4y-7-

Question Number 128611 by malwan last updated on 08/Jan/21

w r i t e t h e e q u a t i o n o f t h e

c i r c l e w h i c h c o n t a i n i n g

t h e p o i n t s (4, - 3), (1, - 2)

a n d c e n t e r o n l i n e

3 x + 4 y = 7

Answered by mr W last updated on 09/Jan/21

(\frac{4 + 1}{2}, \frac{- 3 - 2}{2}) = (\frac{5}{2}, - \frac{5}{2})

\frac{- 3 - (- 2)}{4 - 1} = - \frac{1}{3}

y = - \frac{5}{2} + 3 (x - \frac{5}{2})

- \frac{5}{2} + 3 (x - \frac{5}{2}) = \frac{7}{4} - \frac{3}{4} x

\Rightarrow x = \frac{47}{15}

\Rightarrow y = - \frac{5}{2} + 3 (\frac{47}{15} - \frac{5}{2}) = - \frac{3}{5}

\Rightarrow c e n t e r (\frac{47}{15}, - \frac{3}{5})

{r a d i u s}^{2} = {(4 - \frac{47}{15})}^{2} + {(- 3 + \frac{3}{5})}^{2} = \frac{293}{45}

\Rightarrow e q n . o f c i r c l e :

{(x - \frac{47}{15})}^{2} + {(y + \frac{3}{5})}^{2} = \frac{293}{45}

Commented by malwan last updated on 09/Jan/21

t h a n k y o u s i r

Answered by liberty last updated on 08/Jan/21

let the center point is P (a, b)

(1) R = \sqrt{{(4 - a)}^{2} + {(- 3 - b)}^{2}}

(2) R = \sqrt{{(1 - a)}^{2} + {(- 2 - b)}^{2}}

(3) R = R \Rightarrow 16 - 8 a + a^{2} + 9 + 6 b + b^{2} = 1 - 2 a + a^{2} + 4 + 4 b + b^{2}

\Rightarrow 25 - 8 a + 6 b = 5 - 2 a + 4 b; - 6 a + 2 b = - 20

- 3 a + b = - 10

(3) 3 a + 4 b = 7

(2) + (3) \Rightarrow 5 b = - 3, {\begin{cases} b = - \frac{3}{5} \\ a = \frac{7 - 4 (- \frac{3}{5})}{3} = \frac{47}{15} \end{cases}

we get P (\frac{47}{15}, - \frac{3}{5}) & R = \sqrt{{(4 - \frac{47}{15})}^{2} + {(- 3 + \frac{3}{5})}^{2}} = \sqrt{(\frac{169}{225}) + (\frac{144}{25})}

R = \sqrt{\frac{1465}{225}} . then eq of the circle :\equiv {(x - \frac{47}{15})}^{2} + {(y + \frac{3}{5})}^{2} = \frac{1465}{225}

Commented by malwan last updated on 09/Jan/21

t h a n k y o u s o m u c h

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