CircleHardBloom L3
Question
Consider the circle $x^{2} + y^{2} - 10x - 6y + 30 = 0$. Let $O$ be the centre of the circle and tangent at $A(7,3)$ and $B(5,1)$ meet at $C$. Let $S = 0$ represents family of circles passing through $A$ and $B$, then:
Options
A.Area of quadrilateral $OACB = 4$
B.the radical axis for the family of circles $S = 0$ is $x + y = 10$
C.the smallest possible circle of the family $S = 0$ is $x^{2} + y^{2} - 12x - 4y + 38 = 0$
D.the coordinates of point $C$ are $(7,1)$
Solution
[Key Concept: With $O$ the centre and tangents at $A,B$ meeting at $C$, $OACB$ is a kite; use $OA \perp \ AC$ and the tangent length.]
Area of square $OACB = 2 \times 2 = 4$
Radical axis of the family $S = 0$ is line $AB$ whose equation $x = y + 4$
Smallest possible circle of family $S = 0$ has $AB$ as diameter given by
$$\begin{matrix} & \ (x - 5)(x - 7) + (y - 1)(y - 3) = 0 \\ \Rightarrow \ & x^{2} + y^{2} - 12x - 4y + 38 = 0 \end{matrix}$$
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