Question

The circle
${x}^{2} + {y}^{2} - 4x - 4y + 4 = 0$
is inscribed in a triangle which has two of its sides along the co-ordinate axes.

The locus of the circumcentre of the triangle is
$x + y - xy + k \sqrt{ {x}^{2} + {y}^{2} } = 0$

Find the value of $k$.


✔️✔️Proper solution needed✔️✔️​

Answer 1

Well, the given equation of the inscribed circle is as follows

x^2 + y^2 - 4x - 4y + 4 = 0

So, this means the centre of the inscribed circle is (2,2) and the radius of circle is 2 units.

Inscribed circle radius (r) = 2 units

It has been told that two of the sides of the triangle are the coordinate axes themselves.

This means the given triangle is a right angled traingle , right angled at The origin itself.

Let the line representing the third side be ax+by = c

This means the intercepts of the x-axis will be c/a and the y-intercept will be c/b

Now, Area of the traingle will be = 1/2*c/a*c/b = c^2/2ab

Now,

Semi-perimeter of the traingle is

c/a+c/b + c√1/a^2+1/b^2

So,

Inradius = Area / semi-perimeter

find a relation in a, b and c.

Then think of a feasible linear equation of the line that satisfies the equation dervied by the relation of the above formula.

Then you'll get the line that represents the third side.

Remember, the traingle will be formed in the first quadrant only.

Once, you've got the linear equation then find intercepts and you've also got the lengths of the sides of the traingle.

Find circumradius using relation

Area = (first side)(second side)(third side)/4R

Then, find the coordinate of the circumcircle.

And, then find value of k which will satisfy the found coordinates of circumcircle for the given equation in x,y and k.

Hope this helps you.

Answer 2

ANSWER

k=1

Step By Step Explanation

Plz refer to the attachment for diagram.

Here,let the equation of AB be $\dfrac{x}{a} + \dfrac{y}{a} = 1$

Since the line AB touches the circle we have >

$x^{2} + y^{2} - 4x - 4y + 4 =0$

$\implies \dfrac{ \left | \dftac{2}{a} + \dfrac{2}{b} -1\right | } {\sqrt(\dfrac{1}{a^{2} } + \dfrac {1}{b^{2} }) }$ = 0

Since here ,O(0,0) and C(2,2) lie on the same side of the AB ,hence

$\dfrac{2}{a} + \dfrac{2}{b} -1 > 0$

$\implies \dfrac {-( 2b + 2a - ab)}{\sqrt(a^{2} - b^{2}) }$ = 2

$\implies 2a + 2b -ab + 2\sqrt (a^{2} - b^{2}) }$ = 0

-----------------(i)

Also we know that

$\Delta AOB \: is \: a \: right \: angled \: triangle \: with \: angle,O =90\textdegree$

So , h =$\dfrac{a}{2}$

=> 2h = a. --------(ii)

k = $\dfrac{b}{2}$

=> 2k = b. -----------(iii)

Now, from equation (i),(ii) and (iii) we get >

$4h + 4k -4kh + 2[\sqrt(4h^{2} + 4k^{2}] = 0$

$\implies h + k - kh + 2[\sqrt(h^{2} + k^{2}] = 0$

Hence the locus if Point P(h,k) will be ->

$h + k - kh + 2[\sqrt(x^{2} + y^{2}] = 0$

But we know that the locus of circumcenter is given by

$h + k - kh + 2k[\sqrt(x^{2} + y^{2}] = 0$

=> k = 1

Hence, the value of $k$ will be 1.