Question

(a) Find the equation of the circle passing through the points
(0, 0), (6, 0) and (8, 4).
11
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1 and directriy​

Answer 1

SOLUTION

TO DETERMINE

The equation of the circle passing through the points (0, 0), (6, 0) and (8, 4).

EVALUATION

Let the equation of the circle is

$\sf{ {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 \: \: \: - - - - - (1)}$

Now the circle passes through the point (0,0)

So we have from equation 1

$\sf{ {0}^{2} + {0}^{2} + 2g.0 + 2f.0+ c = 0 }$

$\implies \sf{c = 0}$

So equation 1 becomes

$\sf{ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: \: \: - - - - - (2)}$

Now the circle passes through the point (6,0)

From Equation 2 we get

$\sf{ {6}^{2} + {0}^{2} + 2g.6 + 2f.0 = 0 }$

$\sf{ \implies \: 12g = - 36}$

$\sf{ \implies \: g = - 3}$

Again the circle passes through the point (8,4)

So we get from equation 2

$\sf{ {8}^{2} + {4}^{2} + 2g.8 + 2f.4 = 0 }$

$\sf{ \implies \: 64 + 16 +16g + 8f = 0 }$

$\sf{ \implies \: 80+16g + 8f = 0 }$

$\sf{ \implies \: 80 + 16 \times ( - 3) + 8f = 0 }$

$\sf{ \implies \: 80 - 48 + 8f = 0 }$

$\sf{ \implies \: 8f = - 32 }$

$\sf{ \implies \: f = - 4 }$

Hence the required equation of the circle is

$\sf{ {x}^{2} + {y}^{2} - 6x - 8y = 0}$

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Answer 2

Answer:

SOLUTION

TO DETERMINE

The equation of the circle passing through the points (0, 0), (6, 0) and (8, 4).

EVALUATION

Let the equation of the circle is

\sf{ {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 \: \: \: - - - - - (1)}x

2

+y

2

+2gx+2fy+c=0−−−−−(1)

Now the circle passes through the point (0,0)

So we have from equation 1

\sf{ {0}^{2} + {0}^{2} + 2g.0 + 2f.0+ c = 0 }0

2

+0

2

+2g.0+2f.0+c=0

\implies \sf{c = 0}⟹c=0

So equation 1 becomes

\sf{ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: \: \: - - - - - (2)}x

2

+y

2

+2gx+2fy=0−−−−−(2)

Now the circle passes through the point (6,0)

From Equation 2 we get

\sf{ {6}^{2} + {0}^{2} + 2g.6 + 2f.0 = 0 }6

2

+0

2

+2g.6+2f.0=0

\sf{ \implies \: 12g = - 36}⟹12g=−36

\sf{ \implies \: g = - 3}⟹g=−3

Again the circle passes through the point (8,4)

So we get from equation 2

\sf{ {8}^{2} + {4}^{2} + 2g.8 + 2f.4 = 0 }8

2

+4

2

+2g.8+2f.4=0

\sf{ \implies \: 64 + 16 +16g + 8f = 0 }⟹64+16+16g+8f=0

\sf{ \implies \: 80+16g + 8f = 0 }⟹80+16g+8f=0

\sf{ \implies \: 80 + 16 \times ( - 3) + 8f = 0 }⟹80+16×(−3)+8f=0

\sf{ \implies \: 80 - 48 + 8f = 0 }⟹80−48+8f=0

\sf{ \implies \: 8f = - 32 }⟹8f=−32

\sf{ \implies \: f = - 4 }⟹f=−4

Hence the required equation of the circle is

\sf{ {x}^{2} + {y}^{2} - 6x - 8y = 0}x

2

+y

2

−6x−8y=0

━━━━━━━━━━━━━━━━

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