The problem you’ve provided discusses two circles and asks to find the equation of the tangent $CBD$. Let's break it down based on the information given:

Information:

1. First circle: Center $A(-1, 4)$, passes through point $B(3, 2)$.
2. Second circle: Center $B(3, 2)$, passes through point $A(-1, 4)$.
3. The tangent at $B$ to the first circle intersects the second circle at points $C$ and $D$.

(a) To find the equation of the tangent $CBD$:

We know that the tangent at point $B$ is perpendicular to the radius at $B$, so we need to:

1. Find the slope of the radius from $A$ to $B$. This is the slope of the line passing through $A(-1, 4)$ and $B(3, 2)$: $m_{\text{AB}} = \frac{2 - 4}{3 - (-1)} = \frac{-2}{4} = -\frac{1}{2}$
2. The slope of the tangent will be the negative reciprocal of the slope of $AB$, because the tangent is perpendicular to the radius: $m_{\text{tangent}} = 2$
3. Now, using the point-slope form of a line equation $y - y_1 = m(x - x_1)$, where the point is $B(3, 2)$ and the slope is 2: $y - 2 = 2(x - 3)$ Simplifying: $y - 2 = 2x - 6$ $y = 2x - 4$ Thus, the equation of the tangent $CBD$ is: $y = 2x - 4$

Would you like a detailed breakdown of the next part or have any questions on this?

Follow-up Questions:

1. How do we find the radius of a circle given two points?
2. What is the general equation of a circle?
3. How do we determine if a line is tangent to a circle geometrically?
4. What is the significance of the point of tangency in this context?
5. How can we calculate the coordinates of $C$ and $D$?

Tip:

Remember, the slope of a tangent to a circle is always the negative reciprocal of the slope of the radius at the point of tangency!