To solve these problems, we need to follow these steps:
### (a) Determine the equation of the Circle
To find the equation of the circle with the center at E and passing through points B and T, we need to know the coordinates of E, B, and T. For the purposes of this explanation, let's assume that E has coordinates (e₁, e₂), B has coordinates (b₁, b₂), and T has coordinates (t₁, t₂). However, since you gave an equation in the form (x-a)²+(y-b)² = r², which implies that the radius squared (r²) is 2, we actually only need the coordinates of point E to complete the equation. Let's assume the coordinates of E are (a, b), then the general equation of the circle becomes:
(x-a)² + (y-b)² = 2
Given that the equation provided directly tells us that the radius squared is 2, we don't need the coordinates of B and T to write out the equation. We will only need those coordinates if we want to verify that B and T lie on the circle.
### (b) Determine the equation of the Tangent at point B
For a tangent to the circle at point B(k, k), we need to know the gradient of the radius to point B since the tangent is perpendicular to the radius at the point of tangency.
We have the center of the circle at E(a, b) and the point of tangency at B(k, k).
1. First, calculate the gradient of the line connecting E and B:
[tex]\[ m_{EB} = \frac{k - b}{k - a} \][/tex]
2. The gradient of the tangent line, [tex]\( m_{tangent} \)[/tex], will be the negative reciprocal of [tex]\( m_{EB} \)[/tex] because a tangent is perpendicular to the radius:
[tex]\[ m_{tangent} = -\frac{1}{m_{EB}} = -\frac{k - a}{k - b} \][/tex]
3. Now that we have the gradient of the tangent, we can use the point-slope form to write the equation of the tangent line. With a point (k, k) on the line and [tex]\( m_{tangent} \)[/tex] as the slope, the equation is:
[tex]\[ y - k = m_{tangent}(x - k) \][/tex]
[tex]\[ y - k = -\frac{k - a}{k - b}(x - k) \][/tex]
4. Finally, to explicitly solve this, you need the actual values of a and b from the coordinates of point E. Since we don’t have the specific values, the equation stays in this general form.
Now, if you input the actual numerical values of a, b, and k and simplify the equation, you would get an equation in the form y = mx + c, where m would be the gradient and c would be the y-intercept of the tangent line.
