The equation of a circle with the origin as the centre and passing through the vertices of an equilateral triangle whose altitude is of length 3 units is

Question

The equation of a circle with the origin as the centre and passing through the vertices of an equilateral triangle whose altitude is of length 3 units is (A) x2+y2=9 (B) x2+y2=16 (C) x2+y2=4 (D) x2+y2=1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-xdy1mawxvbye.png" /> Circle circumscribes the triangle ABC and has center at (0 , 0) . AD=3 ∴ OA=(2/3)× 3=2 ∴ radius of circle, OA=2 Thus, equation of circle is x2+y2=4

Q. The equation of a circle with the origin as the centre and passing through the vertices of an equilateral triangle whose altitude is of length $3$ units is

$x^{2} + y^{2} = 9$

$x^{2} + y^{2} = 16$

$x^{2} + y^{2} = 4$

$x^{2} + y^{2} = 1$

Circle circumscribes the triangle $A BC$ and has center at $(0, 0)$ .
$A D = 3$
$∴ O A = \frac{2}{3} \times 3 = 2$
$∴$ radius of circle, $O A = 2$
Thus, equation of circle is $x^{2} + y^{2} = 4$

Q. The equation of a circle with the origin as the centre and passing through the vertices of an equilateral triangle whose altitude is of length 3 units is

x2+y2=9

x2+y2=16

x2+y2=4

x2+y2=1