Question 1 of 3
Consider the equation for circle G.
+y2-28x-10y+205=0
Complete the statement by filling in the blank.
The radius of the circle G is (14,5)
with a center at



Answer :

The radius of a circle is not a coordinate pair, but rather a single non-negative number that represents the distance from the center of the circle to any point on the circle. Therefore, it seems there might be a misunderstanding in the phrasing of the blank in the statement. To find the radius of circle G, however, we first need to identify the standard form of the circle's equation and then determine its center and radius from that standard form. The standard form of the equation of a circle is: \[(x - h)^2 + (y - k)^2 = r^2\] where \( (h, k) \) is the center of the circle and \( r \) is its radius. Given the equation: \[ +y^2 - 28x - 10y + 205 = 0 \] Rearrange the terms to group \( x \) and \( y \): \[ x^2 - 28x + y^2 - 10y = -205 \] Now, we need to complete the square for both \( x \) and \( y \) terms. For \( x \): \[ x^2 - 28x \] Take half of the coefficient of \( x \), which is \(-28\), and square it: \[ \left(\frac{-28}{2}\right)^2 = (-14)^2 = 196 \] Add and subtract this value to complete the square: \[ x^2 - 28x + 196 - 196 \] Which becomes: \[ (x - 14)^2 - 196 \] Now, for \( y \): \[ y^2 - 10y \] Take half of the coefficient of \( y \), which is \(-10\), and square it: \[ \left(\frac{-10}{2}\right)^2 = (-5)^2 = 25 \] Add and subtract this value to complete the square: \[ y^2 - 10y + 25 - 25 \] Which becomes: \[ (y - 5)^2 - 25 \] Putting it all together: \[ (x - 14)^2 - 196 + (y - 5)^2 - 25 = -205 \] Add the constants on the left to balance the equation: \[ (x - 14)^2 + (y - 5)^2 = 196 + 25 - 205 \] \[ (x - 14)^2 + (y - 5)^2 = 16 \] So, the center \((h, k)\) of circle G is \((14, 5)\). The radius \( r \) is the square root of the value on the right-hand side of the equation: \[ r = \sqrt{16} = 4 \] Now we can correctly fill in the blanks: The radius of the circle G is \( 4 \) with a center at \( (14,5) \).

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