Question

A circle of radius $3cm$ can be drawn through two points $A,B$ such that $AB = 6cm$.
A). False
B). True
C). Cannot be determined
D). None of the above

Answer 1

To check if the circle having radius $3cm$ can or cannot be drawn through two points $A,B$ such that $AB = 6cm$, let’s assume that point $A$ is the origin $\left( {0,0} \right)$. Then, we have point $B\left( {6,0} \right)$. We will use these two points now to find the equation of this circle and the centre of the circle. If the distance between the circle and its centre comes out to be $3cm$, then it means that the given statement will be true.
Formula Used:
General equation of a circle passing through a point $P\left( {x,y} \right)$ is given in the form, ${x^2} + {y^2} + 2gx + 2fy + c = 0$ where, $\left( { - g, - f} \right)$ is the centre of the circle, $$$$

Complete step-by-step solution:
Two points are given $A,B$ such that $AB = 6cm$
Let’s assume that point $A$ is the origin $\left( {0,0} \right)$.
$\Rightarrow$ We have point $B\left( {6,0} \right)$.
General equation of a circle passing through a point $P\left( {x,y} \right)$ is given in the form, ${x^2} + {y^2} + 2gx + 2fy + c = 0$ $........\left( 1 \right)$
where,$\left( { - g, - f} \right)$ is the centre of the circle,
It means that points $A,B$ must satisfy the general equation of the circle.
Putting the value of Point $A\left( {0,0} \right)$ in equation $\left( 1 \right)$
${\left( 0 \right)^2} + {\left( 0 \right)^2} + 2g\left( 0 \right) + 2f\left( 0 \right) + c = 0$
$\Rightarrow c = 0$ $......\left( 2 \right)$
Now, Put the value of point $B\left( {6,0} \right)$ in equation $\left( 1 \right)$,
${\left( 6 \right)^2} + {\left( 0 \right)^2} + 2g\left( 6 \right) + 2f\left( 0 \right) + c = 0$
$\Rightarrow 36 + 12g + c = 0$ $......\left( 3 \right)$
Put the value of $c$ from equation $\left( 2 \right)$
$12g = - 36$
$\Rightarrow g = - 3$
From equation $\left( 2 \right)$&$\left( 3 \right)$, we get,
$g = - 3,f = 0$
Then, Point of centre of the circle is $\left( { - g, - f} \right)$
$\Rightarrow \left( {3,0} \right)$ is the centre of the circle.
Clearly, we can see that the distance between centre of the circle $\left( {3,0} \right)$ and point $A\left( {0,0} \right)$ is $3$ which is the radius of the circle.
Then, $AB$ becomes the diameter of the circle. It means that Circle having radius can be drawn through two points $A,B$ such that $AB = 6cm$
Hence, we can say that the given statement is (B) true.

Note: Alternatively, this question can directly be solved observing the radius and difference between two given points. As the radius is given $3cm$ ,
Then the diameter of this circle will be $6cm$ which is the length of $AB$. Clearly, it means we can draw a circle of radius $3cm$ passing through two points $A,B$ such that $AB = 6cm$.