Question

The radius of the circle with centre at the origin is $10$ units. Write the coordinates of the point where the circle intersects the axes. Find the distances between any two of such points.

• A. $Co-ordinates=(10,0)(-10,0)(0,10)(0,-10)$, Distance$=20,10\sqrt{2}$units
• B. $Co-ordinates=(10,0)(-10,0)(0,10)(0,-10)$ Distance$=10,10\sqrt{2}$units
• C. $Co-ordinates=(10,0)(-10,0)(0,10)(0,-10),0$ Distance$=20\sqrt{2}or10\sqrt{2}$units
• D. none

Answer 1

The correct option is A $Co-ordinates=(10,0)(-10,0)(0,10)(0,-10)$, Distance$=20,10\sqrt{2}$units

The equation of circle with radius 10 and centre (0,0)
$x^2+y^2={10}^2$ .................eq(i)
and equation of axes are $x=0$ and $y=0$ .............eq(ii)
putting $x=0$ in eq(i) we get
$\Rightarrow y^2=100$
$\Rightarrow y=\pm 10$
thus we get two coordinates $A(0,10)$ and $C(0,-10)$

putting $y=0$ in eq(i) we get

$\Rightarrow x^2=100$

$\Rightarrow x=\pm 10$

thus we get two coordinates $B(10,0)$ and $D(-10,0)$
Calculation of distances between any two adjacent points
$AB=\sqrt{(0-10{)}^2+(10-0{)}^2}=\sqrt{200}=10\sqrt{2}$units
$AB=BC=CD=DA=10\sqrt{2}$units
and,
Calculation of distances between any two opposite points
$BD=\sqrt{(10-(-10){)}^2+(0-0{)}^2}=20$units
$AC=BD=20$units