The standard equation of a circle with center (h, k) and radius r is given by (x-h)² + (y-k)² = r². We need to transform the given equation into this standard form to identify the center and radius.

The standard equation of a circle with center (h, k) and radius r is given by (x-h)² + (y-k)² = r². We need to transform the given equation into this standard form to identify the center and radius.

By comparing (x+4)² + (y+5)² = 49 with the standard form (x-h)² + (y-k)² = r², we can identify the center (h, k) and the radius r. Here, h = -4, k = -5, and r² = 49, so r = √49 = 7.

Find the centre and the radius of the circle $x^{2}+y^{2}+8 x+10 y-8=0$ .

Answer: The center of the circle is $(-4, -5)$ and the radius is $7$ .

Step-by-step solution

Step 1: Recall the Standard Equation of a Circle

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$ . We need to transform the given equation into this standard form to identify the center and radius.

Step 2: Rearrange and Group Terms

First, rearrange the terms of the given equation, grouping the $x$ terms together and the $y$ terms together, and moving the constant term to the right side of the equation.

Step 3: Complete the Square for x-terms

To complete the square for the $x$ -terms, take half of the coefficient of $x$ (which is $8$ ), square it, and add it to both sides of the equation. Half of $8$ is $4$ , and $4^2$ is $16$ .

Step 4: Complete the Square for y-terms

Similarly, for the $y$ -terms, take half of the coefficient of $y$ (which is $10$ ), square it, and add it to both sides of the equation. Half of $10$ is $5$ , and $5^2$ is $25$ .

Step 5: Rewrite in Standard Form

Now, substitute the completed squares back into the equation and sum the constants on the right side. This gives the equation in the standard form $(x-h)^2 + (y-k)^2 = r^2$ .

Step 6: Identify Center and Radius

By comparing $(x+4)^2 + (y+5)^2 = 49$ with the standard form $(x-h)^2 + (y-k)^2 = r^2$ , we can identify the center $(h, k)$ and the radius $r$ . Here, $h = -4$ , $k = -5$ , and $r^2 = 49$ , so $r = \sqrt{49} = 7$ .

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Find the centre and the radius of the circle x2+y2+8x+10y−8=0x^{2}+y^{2}+8 x+10 y-8=0x2+y2+8x+10y−8=0.