Question

The equation of largest circle passing through the points (1,1) & (2,2) and always in the first quadrant is
a)${{x}^{2}}+{{y}^{2}}-4x-2y+4=0$
b)${{x}^{2}}+{{y}^{2}}-2x-4y+4=0$
c)${{x}^{2}}+{{y}^{2}}-3x-3y+4=0$
d)None

Answer 1

Hint: If you have two points given, then the largest circle can be made by considering those as extra entities of the circle and then use formula \[{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}\] where (a, b) is centre and ‘r’ is radius.

Complete step-by-step answer:

Now let's assume that the centre of the required circle is C(h,k) which passes through P(1,1) and Q(2,2).

Then,

The radius of the circle = CP = CQ =perpendicular distance from C to x axis (CD)

Now,

$CP=CQ=CD$

$C{{P}^{2}}=C{{Q}^{2}}={{(CD)}^{2}}$

${{(h-1)}^{2}}+{{(y-1)}^{2}}={{(x-2)}^{2}}+{{(y-2)}^{2}}={{K}^{2}}$

${{h}^{2}}-2h+1+{{k}^{2}}-2k+1={{h}^{2}}-4h+4+{{k}^{2}}-4k+4={{k}^{2}}$

${{h}^{2}}-2h+1+{{k}^{2}}-2k+1=h-4h+4+{{k}^{2}}-4k+4$

$2h+2k=6\text{ }or\text{ }h+k=3..........(i)$

${{h}^{2}}-2h+1+{{k}^{2}}-2k+1={{k}^{2}}$

${{h}^{2}}-2h+2-2k=0,$

Now putting $k=3-h$ in eq. (i)

So, the equation will be:

${{h}^{2}}-2h+2-2(3-h)=0$

${{h}^{2}}-2h+2-6+2h=0$

${{h}^{2}}=4\Rightarrow h=2$

$k=3-h=3-2=1$

$C(2,1)\text{ and }r=k=1$

So, therefore the equation of circle is:-


${{(x-2)}^{2}}+{{(y-1)}^{2}}={{(1)}^{2}}$

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

Hence, option (a) is the correct answer.

This circle is completely in the 1st quadrant and touches the x-axis.


Note: There can be confusion about how we can find the y-axis, so we can similarly find the equation of another circle which is also completely in the 1st quadrant whose centre C(1,2) ,radius is 1 and touches the y-axis .

${{(x-2)}^{2}}+{{(y-1)}^{2}}={{(1)}^{2}}$