Two question on equation of locus?

1.
A = (0, 0)
Then B = (8, 0)

Consider ΔABC :
Since ∠CAB = ∠CBA
Then, ΔABC is an isosceles triangle, CA = CB

Let (x, y) be the coordinates of C.
CA =CB
√[(x - 0)² + (y - 0)²] = √[(x - 8)² + (y - 0)²]
(x - 0)² + (y - 0)² = (x - 8)² + (y - 0)²
x² + y² = x² - 16x + 64 + y²
16x = 64
x = 4

Locus of C : x = 4


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2.
(a)
Refer to the diagram below.
Draw PA ⊥ OA and cut OA at A.
Join OP.

OA = (26 - 6 × 3) cm = 8 cm

In right-angled ΔPAO :
OP² = PA² + OA²
OP² = (6² + 8²) cm²
OP = 10 cm

If the wire is pivoted at O and rotates in one complete revolution, OP is constant equal to 10 cm.
Then, the locus of P is the circle with center O (0, 0) and radius 10 cm.

(b)
Let (x, y) be the coordinates of P.

OP = 10
OP² = 100
(√[(x - 0)² + (y - 0)²])² = 100
x² + y² = 100

The locus of P : x² + y² = 100

In the first problem, the locus of C is the perpendicular bisector of line segment AB. No coordinates are given for A or B. Perhaps they were cropped from your photo.

The second question could have used some more point labels. Starting from P, and continuing through the length of the wire, let the endpoints and vertices be P, Q, R, O. Construct a line through P, parallel to QR, and let it meet RO at S. Now you should see this:

PQ = QR = RS = 6 cm

Subtracting those from the length of the wire, SO must be 8 cm. That makes ∆PSO a right triangle, and hypotenuse PO = 10 cm.

Presumably the wire remains rigid as it spins about point O, so point P traces a circle of radius 10 cm, centered on O. Again, no coordinates are given, so the equation of the circle is up for grabs. If O is the origin, then the equation you were given makes sense.

1.
https://gyazo.com/046df73fe041ae582b7d5afa3d51242b