Question

Write the equation in standard form of the circle whose center is at (0, 0) and that is tangent to x + y = 6

Answer 1

Answer:

$x^2 + y^2 =18$

Step-by-step explanation:

The standard equation of a circumference has the following formula.

$(x-h) ^ 2 + (y-k) ^ 2 = r ^ 2$

Where the point (h, k) is the center of the circle and r is the radius.

If in this case we know that the circle has center at point (0,0), then its equation will have the following form

$x ^ 2 + y ^ 2 = r ^ 2$

The radius of the circumference will be the distance from the center of the circumference to the point where the circumference is tangent to the line $Ax + Bx + C = 0$

The radio is:

$r=\frac{|Ah + Bk +C|}{\sqrt{A^2+B^2}}$

In this case, the line is

$x + y = 6$

And the center of the circumference is (0, 0)

So

$A = 1\\B = 1\\C = -6\\h = 0\\k = 0$

The radio is:

$r=\frac{|1*0 + 1*0 -6|}{\sqrt{1^2+1^2}}\\\\r=\frac{|-6|}{\sqrt{1^2+1^2}}\\\\r=\frac{6}{\sqrt{2}}$

Finally the equation of the circumference is:

$x^2 + y^2 =(\frac{6}{\sqrt{2}})^2\\\\x^2 + y^2 =18$