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Dec 21, 2014

Find the coordinates of the circumcentre of the triangle whose vertices are (8,6) ,(8,-2) and (2,-2). Also find its circum-radius.

Syeda

Answer.

Let the coordinates of the circumcentre of the triangle be (x, y).

Circumcentre of a triangle is equidistant from each of the vertices.

Distance between (8, 6) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 8)^{2} + (y - 6)^{2}] = âˆš[(x - 8)^{2} + (y + 2)^{2}]

[(x - 8)^{2} + (y - 6)^{2}] = [(x - 8)^{2} + (y + 2)^{2}]

(y - 6)^{2 }= (y + 2)^{2}

y^{2} + 36 - 12y = y^{2} + 4y + 4

36 - 12y = 4y + 4

16y = 32

y = 2

Distance between (2, -2) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 2)^{2} + (y + 2)^{2}] = âˆš[(x - 8)^{2} + (y + 2)^{2}]

[(x - 2)^{2} + (y + 2)^{2}] = [(x - 8)^{2} + (y + 2)^{2}]

(x - 2)^{2 }= (x - 8)^{2}

x^{2} + 4 - 4x = x^{2} - 16x + 64

4 - 4x = -16x + 64

12x = 60

x = 5.

Hence, the coordiantes of the circumcentre of the triangle are (5, 2).

Circumradius = âˆš[(5 - 8)^{2} + (2 - 6)^{2}]

= âˆš(9 + 16)

= âˆš25

= 5 units.

Let the coordinates of the circumcentre of the triangle be (x, y).

Circumcentre of a triangle is equidistant from each of the vertices.

Distance between (8, 6) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 8)

[(x - 8)

(y - 6)

y

36 - 12y = 4y + 4

16y = 32

y = 2

Distance between (2, -2) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 2)

[(x - 2)

(x - 2)

x

4 - 4x = -16x + 64

12x = 60

x = 5.

Hence, the coordiantes of the circumcentre of the triangle are (5, 2).

Circumradius = âˆš[(5 - 8)

= âˆš(9 + 16)

= âˆš25

= 5 units.

Ramesh

Sol:

Let the coordinates of the circumcentre of the triangle be (x, y).

Circumcentre of a triangle is equidistant from each of the vertices.

Distance between (8, 6) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 8)^{2} + (y - 6)^{2}] = âˆš[(x - 8)^{2} + (y + 2)^{2}]

[(x - 8)^{2} + (y - 6)^{2}] = [(x - 8)^{2} + (y + 2)^{2}]

(y - 6)^{2 }= (y + 2)^{2}

y^{2} + 36 - 12y = y^{2} + 4y + 4

36 - 12y = 4y + 4

16y = 32

y = 2

Distance between (2, -2) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 2)^{2} + (y + 2)^{2}] = âˆš[(x - 8)^{2} + (y + 2)^{2}]

[(x - 2)^{2} + (y + 2)^{2}] = [(x - 8)^{2} + (y + 2)^{2}]

(x - 2)^{2 }= (x - 8)^{2}

x^{2} + 4 - 4x = x^{2} - 16x + 64

4 - 4x = -16x + 64

12x = 60

x = 5.

Hence, the coordiantes of the circumcentre of the triangle are (5, 2).

Circumradius = âˆš[(5 - 8)^{2} + (2 - 6)^{2}]

= âˆš(9 + 16)

= âˆš25

= 5 units.

Let the coordinates of the circumcentre of the triangle be (x, y).

Circumcentre of a triangle is equidistant from each of the vertices.

Distance between (8, 6) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 8)

[(x - 8)

(y - 6)

y

36 - 12y = 4y + 4

16y = 32

y = 2

Distance between (2, -2) and (x, y) = Distance between (8, -2) and (x, y)

âˆš[(x - 2)

[(x - 2)

(x - 2)

x

4 - 4x = -16x + 64

12x = 60

x = 5.

Hence, the coordiantes of the circumcentre of the triangle are (5, 2).

Circumradius = âˆš[(5 - 8)

= âˆš(9 + 16)

= âˆš25

= 5 units.