Question

Find the radius of the circle that has its center at (0, −4) and passes through (√13,2)

Answer 1

Given is the center and a point through which a circle passes, find its equation.

Explanation:

• Now we know the general equation of a circle is given by, $->(x-a)^2+(y-b)^2=r^2$    
• Here, the center of the circle is the point $(a,b)$ and the radius of the circle is $r$.
• Now here we have the center of the circle, $a=0,\ \ b=-4$
• Hence, the equation can be given as    $x^2+(y+4)^2=r^2$  
• Now given is the point $(\sqrt{13},2)$ this circle passes through.
• Equating the point in the above equation of circle we get,  
• $->r^2=\sqrt{13} ^2+(2+4)^2\\->r=\sqrt{13+36}\\ ->r=\sqrt{49} \\->r=7\ units$  
• The final equation of the circle is $x^2+y^2+8y-33=0$

Answer 2

Given:

We have given the centre of circle is (0,-4) and the passing point of the circle(√13,2).

To Find:

We have to find the radius of circle?

Step-by-step explanation:

• We know the equation of circle with centre (h,k) and radius r is the given by the equation written below

         $(x-h)^2+(y-k)^2=r^2$

• Now we have given the centre of the circle is at (0,-4) hence we get the value of h is 0 and the value of k is -4.

        $(x-0)^2+(y-(-4))^2=r^2$

• Now simplify the above equation by opening brackets and using the formula of whole square

         $x^2+(y+4)^2=r^2\\x^2+y^2+16+8y=r^2$

• We have used the formula of whole square above which is $(a+b)^2=a^2+b^2+2ab$.

• Now we have given the equation goes through (√13,2) put the value x=√13 and y = 2 in above equation

        $(\sqrt{13} )^2+2^2+16+8(2)=r^2\\13+4+16+16=r^2\\49=r^2\\r=\sqrt{49}=7$

Hence, the radius of circle is 7.