Maths M.C.Q (1 Marks)

The line $4x + 3y = 6$ cuts the coordinate axes at $\Big(\frac{3}{2},\ 0\Big)$ and $(0, 2)$
The coordinates of the incentre is $\Big(\frac{\text{ax}_1+\text{bx}_2+\text{cx}_3}{\text{a+b+c}},\ \frac{\text{ay}_1+\text{by}_2+\text{cy}}{\text{a+b+c}}\Big)$
Here, $\text{a}=\frac{5}{2},\ \text{b}=\frac{3}{2},\ \text{c}=2,\ \text{x}_1=0,\ \text{y}_1\\=0,\ \text{x}_2=0,\ \text{y}_2=2,\ \text{x}_3=\frac{3}{2},\ \text{y}_3=0$
Thus, the coordinates of the incentre:
$\Big(\frac{0+0+3}{6},\ \frac{0+3+0}{6}\Big)$
$=\big(\frac{1}{2},\ \frac{1}{2}\Big)$
The equation of the incircle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\text{a}^2$
Also, radius of the incircle $=\frac{\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}}{\text{s}}$
Here, $\text{s}=\frac{\text{a+b+c}}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$
$\therefore$ Radius of the incircle $=\sqrt{\frac{3(3-\text{a})(3-\text{b}(3-\text{c}))}{3}}$
$=\frac{\sqrt{3\Big(3-\frac{5}{2}\Big)\Big(3-\frac{3}{2}\Big)(3-\text{c})}}{3}$
$=\frac{\sqrt{3\Big(3-\frac{1}{2}\Big)\Big(\frac{3}{2}\Big)}}{3}$
$=\frac{1}{2}$
The equation of circle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{1}{4}$
$\Rightarrow4(\text{x}^2+\text{y}^2)-\text{x}-\text{y}+1=0$

Let the required equation of the circle be $(x - h)^2 + (y - k)^2 = a^2.$
Comparing the given equation $x^2 - y^2 - 2x + 4y - 3 = 0$ with
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0,$ we get:
$a = 1, b = -1, h = 0, g = -1, f = 2, c = -3$
Intersection point $\Big(\frac{\text{hf}-\text{bg}}{\text{ab}-\text{h}^2},\ \frac{\text{gh}-\text{af}}{\text{ab}-\text{h}^2}\Big)=\Big(\frac{-1}{-1},\ \frac{-2}{-1}\Big)=(1,\ 2)$
Thus, the centre of the circle is $(1, 2)$
The equation of the required circle is $(x - 1)^2 + (y - 2)^2 = a^2$
Since circle passes through $(1, 1),$ we have:
$1 = a^2$
$\therefore$ Equation of the required circle:
$(x - 1)^2 + (y - 2)^2 = 1$
$\Rightarrow x^2 + y^2 - 2x - 4y + 4 = 0$

Case $I:$ If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and hasradius a is $x^2 + y^2 - 2ax - 2ay + a^2 = 0.$
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2 + y^2 - 10x - 10y + 25 = 0.$
Case $II:$ If the circle lies in the second quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is $x^2 + y^2 + 2ax - 2ay + a^2 = 0.$
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2 + y^2 + 10x - 10y + 25 = 0.$
Case $III:$ If the circle lies in the third quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is $x^2 + y^2 + 2ax + 2ay + a^2 = 0$
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2 + y^2 + 10x + 10y + 25 = 0.$
Case $IV:$ If the circle lies in the fourth quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is $x^2 + y^2 - 2ax + 2ay + a^2 = 0.$
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2 + y^2 - 10x + 10y + 25 = 0.$
Hence, the required equation of the circle is $x^2+ y^2± 10x ± 10y + 25 = 0.$

Let $\text{ABC}$ be the required equilateral triangle.
The equation of the circle is $x^2 + y^2 - 6x - 8y - 25 = 0.$
Therefore, coordinates of the centre $O$ is $(3, 4).$
Radius of the circle $=\text{OA}=\text{OB}=\text{OC}=\sqrt{9+16+25}=5\sqrt{2}$
In $\Delta\text{BOD},$ we have:
$\sin60^\circ=\frac{\text{DB}}{\text{BO}}$
$\Rightarrow\text{DB}=\frac{\sqrt{3}}{2}(5\sqrt{2})$
$\Rightarrow\text{BC}=2\text{BD}-\sqrt{3}\big(5\sqrt{2}\big)=5\sqrt{6}$
Now, area of $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{BC}^2=\big(5\sqrt{6}\big)^2\\=\frac{\sqrt{3}(150)}{4}=\frac{\sqrt{3}(75)}{2}=\frac{\sqrt{3}(225)}{6}$ square units

The centre of the circle $x^2 + y^2 - 3x + 4y - c = 0$ is $\Big(\frac{3}{2},\ -2\Big).$
Therefore, the centre of the required circle is $\Big(\frac{3}{2},\ -2\Big).$
The equation of the circle is $\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\text{a}^2. \ ......(1)$
Also, circle $(1)$ passes through $(-1, -2).$
$\therefore\Big(-1-\frac{3}{2}\Big)^2+\Big(-2+2\Big)^2=\text{a}^2$
$\Rightarrow\text{a}=\frac{5}{2}$
Substituting the value of a in equation $(1):$
$\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\frac{(2\text{x}-3)^2}{4}+(\text{y}+2)^2=\frac{25}{4}$
$\Rightarrow(2\text{x}-3)^2+4(\text{y}+2)^2=25$
$\Rightarrow\text{x}^2+\text{y}^2-3\text{x}+4\text{y}=0$
Hence, the required equation of the circle is$ x^2 + y^2 - 3x + 4y = 0.$