Write the equation of the circle with center (-2, -3) and tangent to the line 2x + 5y = 10.?

Question

http://i40.photobucket.com/albums/e215/xxs3xp1stolxx/module%20710h/Picture5.png

Answer 1

The distance of the center to the line is d=I -4-15-10)I/sqrt(29)
= sqrt(29)
so your equation is
(x+2)^2+(y+3)^2 = 29

Answer 2

so as to locate the equation of the circle, you will prefer the radius. the common formulation for a circle is: r^2 = (x2 - x1)^2 + (y2 - y1)^2, the place (x1, y1) is the middle. The equation 2x + 5y = 10 is tangential to the circle, so a element that would desire to fulfill the above equation is: (5, 0) So, we've 2 factors, one in all it quite is the middle (-2, -3) and the different is on the fringe (5, 0). So, the radius of the circle could be chanced on applying those 2 factors and the equation of a circle with a center at (-2, -3) with radius, r: r^2 = (x - (-2))^2 + (y - (-3))^2 r^2 = ((5 - (-2))^2)/(0 - (-3))^2 r^2 = 40 9/9 r = 7/3 So, the common equation of this circle is: (7/3)^2 = (x - (-2))^2 + (y - (-3))^2 (7/3)^2 = (x + 2)^2 + (y + 3)^2 40 9/9 = (x + 2)^2 + (y + 3)^2

Answer 3

The equation of circle with center(-2, 3) is

(x +2)^2 + (y + 3)^2 = r^2, where r = radius

the equation of the tangent is

2x + 5y = 10

2x + 5y - 10 --------eqn(1)

the slope of the tangent is

5y = -2x + 10 => y = -2/5 x + 2

so slope is -2/5

the slope of the normal is = -1/(-2/5)

= 5/2

the equation of normal

y - y1 = 5/2(x-x1)

but the normal goes through the center

so y +3 = 5/2(x+2)

2y + 6 = 5x + 10

2y - 5x - 4 = 0 ----------------eqn(2)

the intersection of normal and tangent is point of tangency

solving eqn (1) and eqn(2)

2x + 5y = 10

2y - 5x = 4

multiply (1) with 5 and (2) with 2

10x + 25y = 50

-10x + 4y = 8

add

29 y = 58

y = 2 and

2x+ 10 = 10

2x = 0

x = 0

r = distance between center(-2, -3) and (0, 2)

so radius is sqrt(2^2 + 5^2) = sqrt(29)

r^2 = 29

so equation of circle

(x+2)^2 + (y+3)^2 = 29

Answer 4

top 1

Answer 5

hey they help only for math and they provide 1 free session http://messenger.providesupport.com/messenger/stepsedu.html?ps_s=TSxelDr5EMYF&ps_mht=true and if u want math help they can sure help you just click o this and fill ur name and in question write i want free demo of math as kanpresays