<\/span><\/p>\n <\/span><\/p>\n In the figure shown above, \u201cO\u201d is the circumcentre of \u25b3PQR.<\/span><\/p>\n <\/p>\n <\/p>\n The formula for the distance between any two points \\mathrm{A}\\left(x_{a}, y_{a}\\right) \\text { and } \\mathrm{B}\\left(x_{b}, y_{b}\\right)<\/span> <\/span>is given below:<\/span><\/p>\n \\mathrm{AB}=\\sqrt{\\left(x_{a}-x_{b}\\right)^{2}+\\left(y_{a}-y_{b}\\right)^{2}}<\/span><\/span><\/p>\n In \u25b3PQR, let the coordinates of the vertices P, Q and R be \\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right) \\text { and }\\left(x_{3}, y_{3}\\right)<\/span> <\/span>respectively and the coordinate of the circumcentre O is (a, b).<\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n The distance of all the vertices of the triangle from the circumcentre is equal. We have to follow the following steps to find the coordinates of the circumcentre.<\/span><\/p>\n <\/span><\/p>\n The distance between \\mathrm{P} from \\mathrm{O}=\\mathrm{PO}=\\sqrt{\\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}}<\/span> The distance between Q from O=Q O=\\sqrt{\\left(x_{2}-a\\right)^{2}+\\left(y_{2}-b\\right)^{2}}<\/span><\/span><\/p>\n <\/span><\/p>\n 2. Equating PO = QO\u00a0<\/span><\/p>\n \\sqrt{\\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}}=\\sqrt{\\left(x_{2}-a\\right)^{2}+\\left(y_{2}-b\\right)^{2}}<\/span><\/span><\/p>\n <\/span><\/p>\n \\Rightarrow\\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}=\\left(x_{2}-a\\right)^{2}+\\left(y_{2}-b\\right)^{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Equation PO = RO<\/span><\/p>\n \\sqrt{\\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}}=\\sqrt{\\left(x_{3}-a\\right)^{2}+\\left(y_{3}-b\\right)^{2}}<\/span><\/span><\/p>\n \\Rightarrow\\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}=\\left(x_{3}-a\\right)^{2}+\\left(y_{3}-b\\right)^{2}<\/span><\/span><\/p>\n <\/span><\/p>\n 3. Now we have 2 equations as shown above and 2 unknown variables \u201ca\u201d and \u201cb\u201d<\/span><\/p>\n We have to solve the above two equations to find the coordinate of the circumcentre O (a, b) of \u25b3PQR<\/span><\/p>\n <\/span><\/p>\n <\/b><\/p>\n Given:<\/span><\/p>\n - x_{1}=0<\/span><\/span><\/p>\n <\/span><\/p>\n Let the Coordinates of circumcentre be (a,b)<\/span><\/p>\n Equating the given values in the equations\u00a0<\/span><\/p>\n \\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}=\\left(x_{2}-a\\right)^{2}+\\left(y_{2}-b\\right)^{2} <\/span><\/span><\/p>\n \\Rightarrow \\mathrm{a}^{2}+\\mathrm{b}^{2}=\\mathrm{a}^{2}+81+\\mathrm{b}^{2}-18 \\mathrm{~b} <\/span><\/span><\/p>\n <\/p>\n \\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}=\\left(x_{3}-a\\right)^{2}+\\left(y_{3}-b\\right)^{2} <\/span><\/span><\/p>\n <\/p>\n Hence the circumcentre of \u25b3PQR is (4.5, 4.5)<\/span><\/p>\n <\/span><\/p>\n 2. The circumcentre of <\/span>\u25b3<\/span>PQR is (0,0). If the coordinates of points P and Q are (1,0) and (0,1), find the possible coordinates of point R.<\/span><\/p>\n Given:<\/span><\/p>\n - x_{1}=1<\/span><\/span><\/p>\n Let the coordinate of R be \\left(x_{3}, y_{3}\\right)<\/span><\/span><\/p>\n Equating the given values in the equations\u00a0<\/span><\/p>\n \\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}=\\left(x_{3}-a\\right)^{2}+\\left(y_{3}-b\\right)^{2} <\/span><\/span><\/p>\n \u00a0\\Rightarrow x_{3}^{2}+y_{3}^{2}=1<\/span><\/span><\/p>\n <\/p>\n This shows that the values of x<\/span>3<\/span> and y<\/span>3<\/span> are not constant and can be any value satisfying the equation <\/span>x<\/span>3<\/span>2<\/span>+<\/span>y<\/span>3<\/span>2<\/span>=1<\/span> except the values (1,0) and (0,1) as these are already the other two points of the triangle<\/span><\/p>\n <\/span><\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Circumcentre-of-a-triangle-01-290x300.png\")
<\/span><\/p>\nProperties of circumcentre of the triangle<\/b><\/h2>\n
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Finding circumcentre using the distance formula<\/b><\/h2>\n
![\"\"](\"https:\/\/eistudymaterial.s3.amazonaws.com\/Circumcentre-of-a-triangle-02-300x235.png\")
<\/span><\/p>\n\n
<\/span><\/p>\n
The distance between \\mathrm{R} from \\mathrm{O}=\\mathrm{RO}=\\sqrt{\\left(x_{1}-a\\right)^{2}+\\left(y_{1}-b\\right)^{2}}<\/span><\/span><\/p>\nSolved Examples<\/b><\/h2>\n
\n
- \\mathrm{y}_{1}=0<\/span><\/span><\/p>\n
- x_{2}=0<\/span><\/span><\/p>\n
- \\quad y_{2}=9<\/span><\/span><\/p>\n
- x_{3}=9<\/span><\/span><\/p>\n
- y_{3}=0<\/span><\/span><\/p>\n
\\Rightarrow(0-a)^{2}+(0-b)^{2}=(0-a)^{2}+(9-b)^{2} <\/span>
<\/span><\/p>\n
\\Rightarrow 81-18 \\mathrm{~b}=0 <\/span><\/span><\/p>\n
\\Rightarrow 18 \\mathrm{~b}=81 <\/span><\/span><\/p>\n
=\\Rightarrow \\mathrm{b}=4.5<\/span><\/span><\/p>\n
\u00a0\\Rightarrow(0-a)^{2}+(0-b)^{2}=(9-a)^{2}+(0-b)^{2} <\/span><\/span><\/p>\n
\\Rightarrow \\mathrm{a}^{2}+\\mathrm{b}^{2}=\\mathrm{a}^{2}+81-18 \\mathrm{a}+\\mathrm{b}^{2} <\/span><\/span><\/p>\n
\\Rightarrow 81-18 \\mathrm{a}=0 <\/span><\/span><\/p>\n
\\Rightarrow 18 \\mathrm{~b}=81 <\/span><\/span><\/p>\n
\\Rightarrow \\mathrm{a}=4.5<\/span><\/span><\/p>\n
- \\quad \\mathrm{y}_{1}=0<\/span><\/span><\/p>\n
- x_{2}=0<\/span><\/span><\/p>\n
- \\quad \\mathrm{y}_{2}=1<\/span><\/span><\/p>\n
- \\quad \\mathrm{a}=0<\/span><\/span><\/p>\n
- b=0<\/span><\/span><\/p>\n
\\Rightarrow(1-0)^{2}+(0-0)^{2}=\\left(x_{3}-0\\right)^{2}+\\left(y_{3}-0\\right)^{2} <\/span><\/span><\/p>\n
\\Rightarrow x_{3}^{2}+y_{3}^{2}=1 <\/span><\/span><\/p>\n
\\left(x_{2}-a\\right)^{2}+\\left(y_{2}-b\\right)^{2}=\\left(x_{3}-a\\right)^{2}+\\left(y_{3}-b\\right)^{2} <\/span><\/span><\/p>\n
\\Rightarrow(0-0)^{2}+(1-0)^{2}=\\left(x_{3}-0\\right)^{2}+\\left(y_{3}-0\\right)^{2} <\/span>
<\/span><\/p>\n
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