Q

# 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)

$2\sqrt{2}$

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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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