Question

5 Two tangents to a circle meet at the point

(13,0). The angle between the two tangents
is 46. Find the equation of the circle.
(13.0)

Answer 1

The equation of the circle with center at the origin is
`\(x^2 + y^2 = 169\tan^2(23^\circ)\).`

Let's denote the center of the circle as
`\(O\)` and the point of tangency with one of the tangents as
`\(A\)`. The angle between the two tangents is formed by the radii from the center
`\(O\)` to the points of tangency
`\(A\) and \(B\),` where
`\(B\)` is the point of tangency with the second tangent. The angle between the two tangents is 46 degrees.

The radius at the point of tangency is perpendicular to the tangent. Therefore, the angle
`\(AOB\)` is a right angle. Since the angle between the tangents is given as 46 degrees, the angle
`\(AOA'\)` (where
`\(A'\)` is the reflection of
`\(A\)` across the x-axis) is
`\(46/2 = 23\)` degrees.

Now, we have a right-angled triangle
`\(AOA'\)` where we know the angle
`\(23\)` degrees and the coordinates of
`\(A (13, 0)\)`. Using trigonometry, we can find the length of \(OA\) (the radius) and then use it to find the equation of the circle.

Let
`\(r\)` be the radius, then:

`\[ \tan(23^\circ) = (OA')/(OA) \]`

`\[ \tan(23^\circ) = (0 - r)/(13) \]`

Solving for
`\(r\)`, we get:

`\[ r = -13\tan(23^\circ) \]`

Now, the equation of the circle with center
`\((h, k)\)` is given by:

`\[ (x - h)^2 + (y - k)^2 = r^2 \]`

Substitute
`\( h = 0 \), \( k = 0 \), and \( r = -13\tan(23^\circ) \):`

`\[ x^2 + y^2 = (-13\tan(23^\circ))^2 \]`

Simplify the equation:

`\[ x^2 + y^2 = 169\tan^2(23^\circ) \]`

So, the correct equation of the circle with center at the origin
`\((0, 0)\)` and radius
`\(r = -13\tan(23^\circ)\)` is:

`\[ x^2 + y^2 = 169\tan^2(23^\circ) \]`