
$ A = (2,2),\,\,B = (2,5)\,\,and\,\,C(5,2) $ form a triangle. The circumcentre of $ \Delta ABC $ is
$
A.\,\,\left( {3,3} \right) \\
B.\,\,\left( {2,2} \right) \\
C.\,\,\left( {3.5,3.5} \right) \\
D.\,\,\left( {2.5,2.5} \right) \\
$
Answer
576k+ views
Hint: To find circumcentre we first let centre of a circle as (a, b) then finding radius OA, OB and OC by using distance formula and then equating them in pair to form two equations and then on solving them we get value of ‘a’ and ‘b’ and hence circumcentre of the given triangle.
Distance between two points: $ \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Complete step-by-step answer:
Let (a,b) be the radius of the circle.
Then,
OA = OB
$
\sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {b - 2} \right)}^2}} = \sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {b - 5} \right)}^2}} \\
Sqauring\,\,both\,\,side\,\,to\,\,remove\,\,square\,\,root. \\
\Rightarrow {\left( {a - 2} \right)^2} + {\left( {b - 2} \right)^2} = {\left( {a - 2} \right)^2} + {\left( {b - 5} \right)^2} \\
\Rightarrow {\left( {b - 2} \right)^2} = {\left( {b - 5} \right)^2} \\
\Rightarrow b - 2 = \pm \left( {b - 5} \right) \\
\Rightarrow b - 2 = b - 5\,\,\,\,or\,\,\,\,b - 2 = - b + 5 \\
\Rightarrow 2b = 7 \\
\Rightarrow b = 3.5 \;
$
Also, OB = OC
\[
\sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {b - 5} \right)}^2}} = \sqrt {{{\left( {a - 5} \right)}^2} + {{\left( {b - 2} \right)}^2}} \\
sqauring\,\,both\,\,side\,\, \\
{\left( {a - 2} \right)^2} + {\left( {b - 5} \right)^2} = {\left( {a - 5} \right)^2} + {\left( {b - 2} \right)^2} \\
\]
\[
\Rightarrow {a^2} + 4 - 4a + {b^2} + 25 - 10b = {a^2} + 25 - 10a + {b^2} + 4 - 4b \\
\Rightarrow {{{a^2}}} + {4} - 4a + {{{b^2}}} + {{25}} - 10b - {{{a^2}}} - {{25}} + 10a - {{{b^2}}} - {4} + 4b = 0 \\
\Rightarrow 6a - 6b = 0 \\
\Rightarrow a = b \;
\]
Substituting value of b from above. We have,
$ a = 3.5 $
Hence, from above we see that circumcentre of a triangle is $ \left( {3.5,3.5} \right) $
So, the correct answer is “Option C”.
Note: Circumcentre and Incentre are two different terms as students many times confused with them. Circum-centrre is a centre of the circle formed by side bisectors of a given triangle and in-centre is formed from the angle bisector of the angles of a given triangle.
Distance between two points: $ \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Complete step-by-step answer:
Let (a,b) be the radius of the circle.
Then,
OA = OB
$
\sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {b - 2} \right)}^2}} = \sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {b - 5} \right)}^2}} \\
Sqauring\,\,both\,\,side\,\,to\,\,remove\,\,square\,\,root. \\
\Rightarrow {\left( {a - 2} \right)^2} + {\left( {b - 2} \right)^2} = {\left( {a - 2} \right)^2} + {\left( {b - 5} \right)^2} \\
\Rightarrow {\left( {b - 2} \right)^2} = {\left( {b - 5} \right)^2} \\
\Rightarrow b - 2 = \pm \left( {b - 5} \right) \\
\Rightarrow b - 2 = b - 5\,\,\,\,or\,\,\,\,b - 2 = - b + 5 \\
\Rightarrow 2b = 7 \\
\Rightarrow b = 3.5 \;
$
Also, OB = OC
\[
\sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {b - 5} \right)}^2}} = \sqrt {{{\left( {a - 5} \right)}^2} + {{\left( {b - 2} \right)}^2}} \\
sqauring\,\,both\,\,side\,\, \\
{\left( {a - 2} \right)^2} + {\left( {b - 5} \right)^2} = {\left( {a - 5} \right)^2} + {\left( {b - 2} \right)^2} \\
\]
\[
\Rightarrow {a^2} + 4 - 4a + {b^2} + 25 - 10b = {a^2} + 25 - 10a + {b^2} + 4 - 4b \\
\Rightarrow {{{a^2}}} + {4} - 4a + {{{b^2}}} + {{25}} - 10b - {{{a^2}}} - {{25}} + 10a - {{{b^2}}} - {4} + 4b = 0 \\
\Rightarrow 6a - 6b = 0 \\
\Rightarrow a = b \;
\]
Substituting value of b from above. We have,
$ a = 3.5 $
Hence, from above we see that circumcentre of a triangle is $ \left( {3.5,3.5} \right) $
So, the correct answer is “Option C”.
Note: Circumcentre and Incentre are two different terms as students many times confused with them. Circum-centrre is a centre of the circle formed by side bisectors of a given triangle and in-centre is formed from the angle bisector of the angles of a given triangle.
Recently Updated Pages
Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

The coating formed on the metals such as iron silver class 12 chemistry CBSE

Metals are refined by using different methods Which class 12 chemistry CBSE

What do you understand by denaturation of proteins class 12 chemistry CBSE

Assertion Nitrobenzene is used as a solvent in FriedelCrafts class 12 chemistry CBSE

Trending doubts
Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which places in India experience sunrise first and class 9 social science CBSE

What is pollution? How many types of pollution? Define it

Name 10 Living and Non living things class 9 biology CBSE

What is the full form of pH?


