Answer:
First, let's find the equation of the circle with a centre at (-4,5). the standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the centre of the circle and r is the radius.
So, substituting the center point (-4,5) into the equation, we get:
(x - (-4))^2 + (y - 5)^2 = r^2
(x + 4)^2 + (y - 5)^2 = r^2
Now, let's find the slope of the line 2x + 3y = -19 by putting it into slope-intercept form:
2x + 3y = -19
3y = -2x - 19
y = (-2/3)x - (19/3)
The slope of this line is -2/3.
At a given location, the tangent to a circle is perpendicular to the radius. As a result, we must calculate the radius of the circle with centre (-4,5) that passes through the line's point of tangency (x,y).
The radius of the circle is equal to the length of the perpendicular line segment from the centre to the tangent line (-4,5). This perpendicular line segment will be denoted by the letter d.
We may use the formula for the distance between a point and a line to get d. The distance d between the point (x,y) and the line 2x + 3y = -19 is calculated as follows:
d = |2x + 3y + 19| / sqrt(2^2 + 3^2)
To be tangent to the circle, the radius should be equal to d. Let's call this radius r.
So, we have two equations:
(x + 4)^2 + (y - 5)^2 = r^2 (equation of circle)
d = |2x + 3y + 19| / sqrt(13) (equation of distance between point and line)
Substituting d = r into the second equation, we get:
r = |2x + 3y + 19| / sqrt(13)
We can now substitute this expression for r into the equation of the circle:
(x + 4)^2 + (y - 5)^2 = (|2x + 3y + 19| / sqrt(13))^2
Since the point of tangency lies on the line 2x + 3y = -19, we can substitute (-19 - 3y)/2 for x in the above equation and solve for y:
((-19 - 3y)/2 + 4)^2 + (y - 5)^2 = (|2((-19 - 3y)/2) + 3y + 19| / sqrt(13))^2
Simplifying and solving for y, we get:
y = -5 ± 2√13
Therefore, the two tangent points are (-19/2, -5 + 2√13) and (-19/2, -5 - 2√13).