Question

The line x - 2y + 4 :0 is a tangent to the circle whose centre is the point
A(-1,2).
Find the equation of the radius of the circle.

Answer 1

Let P be the point where the line \(x-2y+4=0\) touches the circle and lets call the coordinates of point P \((x_1,y_1)\).
Next, the equation of a circle of radius r with center at \((x_0,y_0)\) is given by:\[(x-x_0)^2+(y-y_0)^2=r^2\]In your case you are given the center of circle at (-1, 2), therefore the equation of the circle must be:\[(x+1)^2+(y-2)^2=r^2\]The steps you need to follow are:

1. Use implicit differentiation to find the slope of the tangent to this circle at any point

2. You know the line \(x-2y+4=0\) is a tangent to this circle, and we know the slope of this line is 0.5, and we know this line touches the circle at \((x_1,y_1)\), so substitute\(x=x_1\), \(y=y_1\) and slope=0.5 into the differential you found in step 1.

3. This will give you an equation in \(x_1\) and \(y_1\)

4. You know that \(x_1\) and \(y_1\) also lie on the line \(x-2y+4=0\), so substitute \(x=x_1\) and \(y=y_1\) into this equation to get \(y_1\) in terms of \(x_1\)

5. Substitute the expression for \(y_1\) from step 4 into the equation you found in step 3 and solve to find \(x_1\)

6. Use the equation in step 4 to find the corresponding value of \(y_1\)

7. You know that \(x_1\) and \(y_1\) also lie on the circle, so substitute the values \(x_1\) and \(y_1\) into the equation of the circle and solve to find r