Find the equation of the largest circle passing through the point $(1, 1)$ and $(2, 2)$ and which does not cross the boundaries of the first quadrant.

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Solution

The general equation for the tangents circle b/w $(1, 1) & (2, 2)$
The center of the circle will be on the perpendicular bisector of $| A B |$ .This perpendicular bisector has equation $x + y = 3$
Since it passes through their mid point $(\frac{3}{2}, \frac{3}{2}) .$
So let the general equation be
$x^{2} + y^{2} - 3 x + 4 + k (x - y) = 0$
When it intersect $x - a x i s, y = 0$
So, we get
$x^{2} - 3 x + 4 + k x = 0 x^{2} - (3 - k) x + 4 = 0$
The equation has real roots, we get
$△ = {(3 - k)}^{2} - 4.4 = 0 9 + k^{2} - 6 k - 16 = 0 k = 6 \pm \sqrt{36 - 28} - 7 < k < 1$
So for $k$ to have non-intersecting axes $k$ has to be $- 1 < k > 1$
Maximum radius achieve $= \pm 1.$
In the equation
$x^{2} + y^{2} - 3 x - 3 y + 4 + k (x - y) = 0 x^{2} + y^{2} + (k - 3) x - (k + 3) y + 4 = 0 r = \sqrt{\frac{{(k - 3)}^{2}}{2} + \frac{(k + 3)^{2}}{2}} - 4$
For $k \pm 1$ we need to have
$r = 1$ as max radius,
Hence, the equation are
Once when center $(2, 1)$
$x^{2} + y^{2} - 4 x - 2 y + 4 = 0 &$
Once when center $(1, 2)$
$x^{2} + y^{2} - 2 x - 2 y + 4 = 0$

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