A circle has the equation x2−12x+27=−y2+8y.
What is the equation of the circle in standard form, the location of its center, and the length of its radius?

The equation of the circle is (x−6)2+(y−4)2=25; the center is at (6,4), and the radius is 5 units.

The equation of the circle is (x−6)2+(y−4)2=5; the center is at (6,4), and the radius is 5–√ units.

The equation of the circle is (x−6)2+(y−4)2=5; the center is at (−6,−4), and the radius is 5–√ units.

The equation of the circle is (x−6)2+(y−4)2=25; the center is at (−6,−4), and the radius is 5 units.

It's the first option.

Step-by-step explanation:

x^2 - 12x + 27 = - y^2 + 8y

x^2 - 12x + y^2 - 8y = -27    Completing the squares:

(x - 6)^2  - 36  + (y - 4)^2 - 16 = -27

(x - 6)^2 + (y - 4)^2 = -27 + 36 + 16

(x- 6)^2 + (y - 4)^2 = 25 is the standard equation.

The center is at (6, 4) and the radius is 5 units.

first option

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² - 12x + 27 = - y² + 8y ( subtract - y² + 8y from both sides )

x² - 12x + y² - 8y + 27 = 0 ( subtract 27 from both sides )

x² - 12x + y² - 8y = - 27

Using the method of completing the square on both the x and y terms

add ( half the coefficient of the x/ y term )² to both sides

x² + 2(- 6)x + 36 + y² + 2(- 4)y + 16 = - 27 + 36 + 16

(x - 6)² + (y - 4)² = 25 ← in standard form

with centre = (6, 4 ) and r² = 25 ⇒ r = = 5 → first option

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.