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Question: How do you find the radius of a circle with the equation \( {x^2} - 8x + {y^2} - 4y - 5 = 0 \) ?...

How do you find the radius of a circle with the equation x2−8x+y2−4y−5=0{x^2} - 8x + {y^2} - 4y - 5 = 0 ?

Explanation

Solution

Hint : In this question, we have to find the radius of a circle from the given equation of circle. (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2} is the standard form of equation of a circle, where (h,k)(h,k) are the coordinates of the centre of the circle and rr is the radius of the circle (the distance between the centre of the circle and any point on the boundary of the circle is known as the radius of the circle). Thus to find the radius of the circle we will convert the given equation to the standard form and then compare both of them.

Complete step by step solution:
We are given that x2−8x+y2−4y−5=0{x^2} - 8x + {y^2} - 4y - 5 = 0
We know that standard form of the equation of a circle is (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
So we will add and subtract some terms in the given question to convert them into standard form –
x2−8x+16−16+y2−4y+4−4−5=0 ⇒x2−8x+(4)2+y2−4y+(2)2−5−4−16=0   {x^2} - 8x + 16 - 16 + {y^2} - 4y + 4 - 4 - 5 = 0 \\\ \Rightarrow {x^2} - 8x + {(4)^2} + {y^2} - 4y + {(2)^2} - 5 - 4 - 16 = 0 \;
We know that a2+b2−2ab=(a−b)2{a^2} + {b^2} - 2ab = {(a - b)^2} and a2+b2+2ab=(a+b)2{a^2} + {b^2} + 2ab = {(a + b)^2} , using these two identities in the obtained equation, we get –
⇒(x−4)2+(y−2)2−25=0 ⇒(x−4)2+(y−2)2=25   \Rightarrow {(x - 4)^2} + {(y - 2)^2} - 25 = 0 \\\ \Rightarrow {(x - 4)^2} + {(y - 2)^2} = 25 \;
Now, on comparing the obtained equation with the standard form, we get –
r2=25 ⇒r=±5   {r^2} = 25 \\\ \Rightarrow r = \pm 5 \;
Length can never be negative, so we reject the negative value.
Hence the radius of a circle with the equation x2−8x+y2−4y−5=0{x^2} - 8x + {y^2} - 4y - 5 = 0 is 55 .
So, the correct answer is “ 55 ”.

Note : After converting the given equation into the standard way, we can also use it to draw its graph as we will get the coordinates of the centre and the radius of the circle. The coordinates of the centre of this circle is (4,2)(4,2) . There are various ways to write the equation of a circle, but the standard way is (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2} .