Find the equations of tangents from point #(-2,11)# to the circle #x^2+y^2=25#?
2 Answers
Please see below.
Explanation:
A quick sketch (or even a mental image) will show us that
The point
We expect two solutions, one tangent to the circle in the second quadrant and one in the first.
Clearly then, we need only consider the upper semicircle whose equation is:
So
The point of tangency is a point on the semicircle. If we call the
The slope of the tangent at that point is
The equation of a line tangent to the semicircle at the point where
We want
Get rid of the fraction by multiplying by
Square both sides and solve the resulting quadratic to get
Recall that
The tangent line through
And the tangent line through
Writing the equation of these two lines is left to the reader.
The points of tangency are not clear in this graph, but you can zoom in and out and drag the graph around in the window.
graph{(x^2+y^2-25)(y-11-24/7(x+2))(y-11+4/3(x+2))=0 [-22.47, 28.9, -9.22, 16.44]}
This picture (using Desmos) may be more clear, but you cannot change the view.
The tangents are
Explanation:
Let the slope of tangent be
Now this cuts the circle
or
or
The above quadratic equation can have up to two solutions. We have two real roots when discriminant is positive and it shows the line cutting the circle at two distinct points. When discriminant is negative, we do not have any real root, which represents the line not touching the circle at all.
And when the line touches the circle, there is only one root (or two coincident roots) of quadratic equation, which happens when the discriminant is equal to zero. As the line we are seeking is a tangent i.e. touches the circle, the discriminant must be zero i.e.
or
i.e.
or
or
Hence
and tangents are
and
graph{(x^2+y^2-25)(24x-7y+125)(4x+3y-25)=0 [-19.92, 20.08, -7.6, 12.4]}