A circle passes through the vertices of a triangle ABC.If the co-ordinates of the orthocenter and the centroid of the triangle are (1,2) and (3,4) and the equation of BC is x-y=0,then the radius of the circle is: 1)sqrt 5 2)2*sqrt 5 3)3*sqrt 5 4)4*sqrt 5

To find the radius of the circumcircle of triangle ABC, we can utilize the relationship between the orthocenter, centroid, and circumradius of a triangle. The circumradius $R$ can be calculated using the formula that relates these points. The formula is given by:

Key Relationships in Triangle Geometry

In any triangle, the following relationship holds:

• Let $G$ be the centroid, $H$ be the orthocenter, and $R$ be the circumradius.
• The distance $GH$ can be expressed as $GH=\frac{R}{3}\sqrt{3}$.

Given the coordinates of the orthocenter $H(1,2)$ and the centroid $G(3,4)$, we can first calculate the distance $GH$.

Calculating the Distance GH

The distance $GH$ can be calculated using the distance formula:

$GH=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Substituting the coordinates of $G$ and $H$:

$GH=\sqrt{(3-1{)}^2+(4-2{)}^2}=\sqrt{(2{)}^2+(2{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

Relating GH to the Circumradius

Now, using the relationship $GH=\frac{R}{3}\sqrt{3}$, we can set up the equation:

$2\sqrt{2}=\frac{R}{3}\sqrt{3}$

To isolate $R$, we multiply both sides by $3$ and divide by $\sqrt{3}$:

$R=2\sqrt{2}\cdot \frac{3}{\sqrt{3}}=2\sqrt{2}\cdot \sqrt{3}=2\sqrt{6}$

Finding the Radius of the Circumcircle

Now, we need to relate $R$ to the options provided. We can express $R$ in terms of the radius options given:

• Option 1: $\sqrt{5}$
• Option 2: $2\sqrt{5}$
• Option 3: $3\sqrt{5}$
• Option 4: $4\sqrt{5}$

Since $R=2\sqrt{6}$, we can approximate $\sqrt{6}$ as approximately $2.45$. Thus, $R$ is approximately $2\times 2.45=4.9$, which is close to $2\sqrt{5}$ since $\sqrt{5}\approx 2.24$ and $2\sqrt{5}\approx 4.48$.

Therefore, the radius of the circumcircle of triangle ABC is:

Final Answer

2√5