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Solve it, The radius of the circle passing through the points (1,2),(5,2) and (5,-2) is

The radius of the circle passing through the points (1,2),(5,2) and (5,-2) is 

  • Option 1)

    5\sqrt{2}

  • Option 2)

    2\sqrt{5}

  • Option 3)

    3\sqrt{2}

  • Option 4)

    2\sqrt{2}

 
Answers (2)
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As we learnt in

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 On substituting the  value of  (x, y) as (1, 2), we get 

(1-h)^{2}+(2-k)^{2}=r^{2}            ....................(1)

 On substituting the  value of  (x, y) as (5, 2), we get

(5-h)^{2}+(2-k)^{2}=r^{2}        .........................(2)

On substituting the  value of  (x, y) as (5, -2), we get

(5-h)^{2}+(-2-k)^{2}=r^{2}       .......................(3)

On subtracting  Eq. (1) - (2), we get 

h = 3

On subtacting Eq. (2) - Eq. (3), we get

k = 0

\therefore r = 2\sqrt{2}

 

 

 

On


Option 1)

5\sqrt{2}

Incorrect

Option 2)

2\sqrt{5}

Incorrect

Option 3)

3\sqrt{2}

Incorrect

Option 4)

2\sqrt{2}

Correct

As we learnt in

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 On substituting the  value of  (x, y) as (1, 2), we get 

(1-h)^{2}+(2-k)^{2}=r^{2}            ....................(1)

 On substituting the  value of  (x, y) as (5, 2), we get

(5-h)^{2}+(2-k)^{2}=r^{2}        .........................(2)

On substituting the  value of  (x, y) as (5, -2), we get

(5-h)^{2}+(-2-k)^{2}=r^{2}       .......................(3)

On subtracting  Eq. (1) - (2), we get 

h = 3

On subtacting Eq. (2) - Eq. (3), we get

k = 0

\therefore r = 2\sqrt{2}

 

 

 

On


Option 1)

5\sqrt{2}

Incorrect

Option 2)

2\sqrt{5}

Incorrect

Option 3)

3\sqrt{2}

Incorrect

Option 4)

2\sqrt{2}

Correct

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