Write down the equation of a circle with a radius of 5 cm, which passes through

Write down the equation of a circle with a radius of 5 cm, which passes through the point (1; 8), and its center is on the bisector of the first coordinate quarter.

The equation of a circle with radius R, centered at a point with coordinates (a; b) in general looks as follows (x – a) ² + (y – b) ² = R², where (x; y) are the coordinates of an arbitrary point of the circle. It is known from the problem statement that the radius of the circle is R = 5 cm, and its center is on the bisector of the first coordinate quarter, that is, a = b and a> 0. In addition, it passes through point A (1; 8). Substitute the values ​​of the indicated quantities into the formula and make the calculations: (1 – a) ² + (8 – a) ² = 5². Let’s open the brackets and give similar terms: а² – 9 ∙ а + 20 = 0. We obtain two solutions of the quadratic equation with respect to the unknown а, these are а₁ = 4 and а₂ = 5. It turns out that the condition of the problem is satisfied by two equations of the circle (x – 4) ² + (y – 4) ² = 5² and (x – 5) ² + (y – 5) ² = 5².
Answer: the equation of the circle with the given parameters has the form (x – 4) ² + (y – 4) ² = 5² or (x – 5) ² + (y – 5) ² = 5².