IIT-JEE 1987Question 1
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 $x+y-xy+k\left(x^{2}+y^{2}\right)^{1/2}=0$. Find $k$.
Circle
Answer$k=1$
Solution
Concept. The triangle is right-angled at $O$ with vertices $(a,0)$, $(0,b)$. Its circumcentre is the midpoint of the hypotenuse; its inradius is $2$. Formulas used. (1) Inradius of a right triangle: $\rho=\frac{a+b-c}{2}$, $c$ the hypotenuse.(2) Circumcentre of a right triangle $=$ midpoint of the hypotenuse; $R=\frac c2$.Step 1 - Given circle. $(x-2)^2+(y-2)^2=4$: centre $(2,2)$, radius $2$. Step 2 - Use (1). $a+b-\sqrt{a^2+b^2}=4$. Squaring $(a+b-4)^2=a^2+b^2$: $$ab-4(a+b)+8=0.$$ Step 3 - Use (2). $(x,y)=\left(\frac a2,\frac b2\right)$. Put $a=2x$, $b=2y$: $$4xy-8(x+y)+8=0\Rightarrow xy=2(x+y)-2.$$ Also $\sqrt{x^2+y^2}=\frac c2=\frac{a+b-4}2=x+y-2$.Step 4 - Substitute in the given locus. $$x+y-xy+k\sqrt{x^2+y^2}=x+y-2(x+y)+2+k(x+y-2)=(k-1)(x+y-2).$$ Step 5 - Result. $x+y-2=\frac c2>0$, so the expression is $0$ only if $k=1$. Answer: $k=1$