Geometrical problems regarding circle and its properties Questions in English

(B) Let the center of the circle be $(h, k)$. Since the circle passes through $(0,0)$, its radius $r$ is $\sqrt{h^2 + k^2}$.
The circle touches the lines $x + y - 1 = 0$ and $x - y - 1 = 0$. Thus, the perpendicular distance from $(h, k)$ to these lines must equal $r$.
$r = \frac{|h + k - 1|}{\sqrt{1^2 + 1^2}} = \frac{|h - k - 1|}{\sqrt{1^2 + (-1)^2}}$.
This implies $|h + k - 1| = |h - k - 1|$.
Case $1$: $h + k - 1 = h - k - 1 \implies 2k = 0 \implies k = 0$.
Substituting $k=0$ into $r^2 = h^2 + k^2$, we get $r^2 = h^2$. Also $r = \frac{|h - 1|}{\sqrt{2}}$, so $h^2 = \frac{(h-1)^2}{2} \implies 2h^2 = h^2 - 2h + 1 \implies h^2 + 2h - 1 = 0$. This gives two values for $h$, so two circles.
Case $2$: $h + k - 1 = -(h - k - 1) \implies h + k - 1 = -h + k + 1 \implies 2h = 2 \implies h = 1$.
Substituting $h=1$ into $r^2 = h^2 + k^2$, we get $r^2 = 1 + k^2$. Also $r = \frac{|1 + k - 1|}{\sqrt{2}} = \frac{|k|}{\sqrt{2}}$, so $r^2 = \frac{k^2}{2}$.
Equating these, $1 + k^2 = \frac{k^2}{2} \implies \frac{k^2}{2} = -1$, which has no real solution for $k$.
Thus, there are $2$ such circles.

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
Since it passes through $(0, 1)$, we have $h^2 + (1 - k)^2 = r^2$. $(1)$
Since it passes through $(1, 1)$, we have $(1 - h)^2 + (1 - k)^2 = r^2$. $(2)$
Equating $(1)$ and $(2)$: $h^2 = (1 - h)^2 \implies h^2 = 1 - 2h + h^2 \implies 2h = 1 \implies h = \frac{1}{2}$.
The slope of the tangent to $y = x^2$ at $(1, 1)$ is $\frac{dy}{dx} = 2x|_{x=1} = 2$. The normal at $(1, 1)$ has slope $-\frac{1}{2}$.
The centre $(h, k)$ lies on the normal line passing through $(1, 1)$ with slope $-\frac{1}{2}$: $\frac{k - 1}{h - 1} = -\frac{1}{2}$.
Substituting $h = \frac{1}{2}$: $\frac{k - 1}{1/2 - 1} = -\frac{1}{2} \implies \frac{k - 1}{-1/2} = -\frac{1}{2} \implies k - 1 = \frac{1}{4} \implies k = \frac{5}{4}$.
Thus, the centre is $(\frac{1}{2}, \frac{5}{4})$.

(B) The distance $OP = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.
In the right-angled triangle $\triangle OAP$, $OA = r$ and $AP = \sqrt{OP^2 - r^2} = \sqrt{100 - r^2}$.
The area of $\triangle OAP = \frac{1}{2} \times OA \times AP = \frac{1}{2} r \sqrt{100 - r^2}$.
The area of quadrilateral $PAOB = 2 \times \text{Area}(\triangle OAP) = r \sqrt{100 - r^2}$.
Let $f(r) = r \sqrt{100 - r^2}$. To maximize $f(r)$, maximize $f(r)^2 = r^2(100 - r^2) = 100r^2 - r^4$.
Let $g(r) = 100r^2 - r^4$. Differentiating with respect to $r$: $g'(r) = 200r - 4r^3$.
Setting $g'(r) = 0$, we get $4r(50 - r^2) = 0$, so $r^2 = 50$, which means $r = \sqrt{50} = 5\sqrt{2}$.
Thus, the area is maximum when $r = 5\sqrt{2}$.

(B) $1$. The equation of the circle is $x^2 + y^2 - 2x - 2y - 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $g = -1, f = -1, c = -2$.
$2$. The center $C$ is $(-g, -f) = (1, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2$.
$3$. The perpendicular distance $d$ from $C(1, 1)$ to the line $x + y - 4 = 0$ is $d = \frac{|1 + 1 - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.
$4$. In $\triangle ABC$, $AC = BC = r = 2$. The height of the triangle from $C$ to chord $AB$ is $d = \sqrt{2}$.
$5$. The length of the chord $AB = 2\sqrt{r^2 - d^2} = 2\sqrt{4 - 2} = 2\sqrt{2}$.
$6$. Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times d = \frac{1}{2} \times 2\sqrt{2} \times \sqrt{2} = 2$.

(D) For circle $C_1: x^2 + y^2 - 6x = 0$, center $C_1 = (3, 0)$ and radius $r_1 = \sqrt{3^2 + 0^2 - 0} = 3$.
For circle $C_2: x^2 + y^2 + 6x + 2y + 1 = 0$, center $C_2 = (-3, -1)$ and radius $r_2 = \sqrt{(-3)^2 + (-1)^2 - 1} = \sqrt{9 + 1 - 1} = 3$.
The distance between the centers $d = \sqrt{(3 - (-3))^2 + (0 - (-1))^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$.
Since $r_1 + r_2 = 3 + 3 = 6$ and $\sqrt{37} > 6$, the distance between the centers is greater than the sum of the radii $(d > r_1 + r_2)$.
Therefore, the circles are separate and do not intersect or touch each other.
In this case, the number of common tangents is $4$.