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

Question

thanks!

Answer 1

The line 2x + 5y = 10 passes through (0,2) and (5,0) and has a slope of -2/5. Graph this line. Graph the point (-2,3) and then then draw a segment having slope 5/2 from (-2,3) to the line 2x + 5y = 10. This is the radius of the circle, which conveniently intersects the line at (0,2). Find the length of this segment using the distance formula, and you'll find that the radius length is the square root of 29. Then use the formula for the equation of a circle to finish this out.

(x - h)^2 + (y - k)^2 = r^2
(x + 2)^2 + (y + 3)^2 = 29

Answer 2

The equation of circle with middle(-2, 3) is (x +2)^2 + (y + 3)^2 = r^2, the place r = radius the equation of the tangent is 2x + 5y = 10 2x + 5y - 10 --------eqn(a million) the slope of the tangent is 5y = -2x + 10 => y = -2/5 x + 2 so slope is -2/5 the slope of the traditional is = -a million/(-2/5) = 5/2 the equation of regular y - y1 = 5/2(x-x1) however the traditional is going by way of the middle so y +3 = 5/2(x+2) 2y + 6 = 5x + 10 2y - 5x - 4 = 0 ----------------eqn(2) the intersection of regular and tangent is factor of tangency fixing eqn (a million) and eqn(2) 2x + 5y = 10 2y - 5x = 4 multiply (a million) with 5 and (2) with 2 10x + 25y = 50 -10x + 4y = 8 upload 29 y = fifty 8 y = 2 and 2x+ 10 = 10 2x = 0 x = 0 r = distance between middle(-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 3

In order to find the equation of the circle, you will need the radius. The general formula for a circle is:

r^2 = (x2 - x1)^2 + (y2 - y1)^2, where (x1, y1) is the center.

The equation 2x + 5y = 10 is tangential to the circle, so a point that would satisfy the above equation is: (5, 0)

So, now we have 2 points, one of which is the center (-2, -3) and the other is on the perimeter (5, 0). So, the radius of the circle can be found using these two points 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 = 49/9
r = 7/3

So, the general equation of this circle is:
(7/3)^2 = (x - (-2))^2 + (y - (-3))^2
(7/3)^2 = (x + 2)^2 + (y + 3)^2
49/9 = (x + 2)^2 + (y + 3)^2

Answer 4

the radium is the distance of(-2-3)to 2x+5y-10=0 which is
abs value of (2*(-2)+5*(-3)-10))/sqrt(4+9) = 29/sqrt(13)
so the equation is
(x+2)^2+(y+3)^2 =841/13
I made a mistake
the sqrt should be sqrt(4+25) = sqrt(29)
so the equation becomes
(x+2)^2+(y+3)^2 =29