equation 2x-3y=10 touches the circle with centre m (-2,4)
the distance from (-2,4) to the line 2x-3y-10 = 0 is 2√13, so that's the radius of the circle.
(x+2)^2 + (y-4)^2 = 52
you can see this at
http://www.wolframalpha.com/input/?i=plot+%28x%2B2%29^2+%2B+%28y-4%29^2+%3D+52+and+2x-3y%3D10
To determine whether the equation 2x-3y=10 touches the circle with center M(-2,4), we need to compare the distance between the center of the circle and the line to the radius of the circle.
The equation of a circle centered at point M(-2,4) can be written in standard form as (x - (-2))^2 + (y - 4)^2 = r^2, where r is the radius of the circle.
First, let's find the distance between the line and the center of the circle. We'll use the formula for the distance between a point (x1, y1) and a line Ax + By + C = 0, which is given by:
d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
For the line 2x-3y=10, A=2, B=-3, C=-10, and the center of the circle is M(-2,4), so x1=-2 and y1=4.
Plugging in the values, we have:
d = |2(-2) + (-3)(4) - 10| / sqrt(2^2 + (-3)^2)
= |-4 - 12 - 10| / sqrt(4 + 9)
= |-26| / sqrt(13)
= 26 / sqrt(13)
Next, we need to find the radius of the circle. Since we don't have any additional information about the circle, we cannot determine the exact value of the radius. However, we can set it as any positive value 'r'.
Now we compare the distance between the line and the center of the circle (26 / sqrt(13)) with the radius 'r'. If the distance is equal to the radius (d = r), then the line touches the circle. If the distance is less than the radius (d < r), then the line intersects the circle. If the distance is greater than the radius (d > r), then the line does not touch or intersect the circle.
Therefore, to conclusively determine whether the line 2x - 3y = 10 touches the circle with center M(-2,4), we need to know the value of the radius 'r' of the circle.