# Me pueden ayudar con estos ejercicios de circunferencia?

1. Una circunferencia tangente a los ejes coordenados tiene su centro en la recta x-y=0 y su radio es 7. ¿Cuál es su ecuación?

2. Halla la ecuación de la circunferencia que pasa por el origen y los puntos (-2,-7) y (3,-5)

### 3 respuestas

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1. The centre line runs through the first and third quadrants, so the equation of the circle can be (x-7)² + (y-7)² = 49 or (x+7)² + (y+7)² = 49.

2. Find the equation of the circumference that passes through the origin and the points (-2, -7) and (3, -5).

2(x₂-x₁)x + 2(y₂-y₁)y = x₂² - x₁² + y₂² - y₁², so the perpendicular bisector of (0,0) and (-2, -7) is 4x + 14y = -53, and the perpendicular bisector of (0,0) and (3, -5) is 3x - 5y = 17.

The point of intersection of ax+by=c and dx+ey=f is ((ce-bf)/(ae-bd), (af-cd)/(ae-bd)) where the lines are not parallel, so these two lines intersect at (-27/62, -227/62).

The equation of the circle passing through the origin and centred at (x₁,y₁) is (x-x₁)²+(y-y₁)² = x₁²+y₁², so the circle centred at (-27/62, -227/62) and passing through (0,0) is (x+27/62)²+(y+227/62)² = 26129/1922.

Personally, I like the first question, but the fractions seem overly complicated in the second question.

• 2. (x-x0)^2 + (y-y0)^2 = r^2 , plug in (x,y)=(0,0), and (x,y)=(-2,-7), and (x,y)= (3,-5). Get after expanding

x0^2+y0^2 = r^2

x0^2+y0^2+4x0+14y0+53 = r^2

x0^2+y0^2-6x0+10y0+34 = r^2

Plug the first equation into the left side of the other two and cancel r^2's, which gives

4x0+14y0+53 = 0 and

-6x0+10y0+34 = 0

Solve and get center is (x0,y0) = (-27/62, -227/62),

plug into first equation and get r^2 = 26129/1922

Hence the equation of the required circle is

(x+27/62)^2 + (y+227/62)^2 = 26129/1922.

For picture see https://www.desmos.com/calculator/va6hzokble

• center at (7,7)

(x-7)² + (y-7)² = 49

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