Circles
Circumcentre and orthocentre
Grade 11
Question:
<p>Let the orthocentre and centroid of a triangle be \(A(-3, 5)\) and \(B(3, 3)\), respectively. If \(C\) is the circumcentre of this triangle, then the radius of the circle having line segment \(AC\) as diameter, is</p>
<p>\(2\sqrt{10}\)</p>
<p>\(3\sqrt{\dfrac{5}{2}}\)</p>
<p>\(\dfrac{3\sqrt{5}}{2}\)</p>
<p>\(\sqrt{10}\)</p>
Step-by-Step Solution
Key Concept: Use Euler's line property: for any triangle, centroid G divides the line segment from circumcenter O to orthocenter H in ratio 1:2, i.e., OG:GH = 1:2. This gives us the circumcenter, then find the radius of circle with AC as diameter.
<p><strong>Step 1:</strong> Apply Euler's line property. If O is circumcenter, G is centroid, and H is orthocenter, then:</p><p>$$\vec{OG}:\vec{GH} = 1:2$$</p><p>This means: $$\vec{G} = \frac{\vec{O} + 2\vec{H}}{3}$$</p><p><strong>Step 2:</strong> Given H = A(-3, 5) and G = B(3, 3). Solve for circumcenter C:</p><p>$$3 = \frac{C_x + 2(-3)}{3} \Rightarrow 9 = C_x - 6 \Rightarrow C_x = 15$$</p><p>$$3 = \frac{C_y + 2(5)}{3} \Rightarrow 9 = C_y + 10 \Rightarrow C_y = -1$$</p><p>So C = (15, -1)</p><p><strong>Step 3:</strong> Find distance AC:</p><p>$$AC = \sqrt{(15-(-3))^2 + (-1-5)^2} = \sqrt{18^2 + (-6)^2} = \sqrt{324 + 36} = \sqrt{360} = 6\sqrt{10}$$</p><p><strong>Step 4:</strong> The radius of circle with AC as diameter:</p><p>$$r = \frac{AC}{2} = \frac{6\sqrt{10}}{2} = 3\sqrt{10}$$</p><p>∴ Answer: A</p>
Correct Answer: A