Question

The equation of the circle which passes through the points $(2,-2)$ and $(3,4)$ and whose centre lies on the line $x+y=2$, is

• A. $(x+0.7{)}^2+(y-1.3{)}^2=13.58$
• B. $(x-0.3{)}^2+(y+0.7{)}^2=12.58$
• C. $(x-0.7{)}^2+(y-1.3{)}^2=12.58$
• D. $(x+0.7{)}^2+(y+1.3{)}^2=13.58$

Answer 1

Answer: (C)

Let the equation of the circle be $(x-h{)}^2+(y-k{)}^2=r^2$.
Since, the circle passes through $(2,-2)$ and $(3,4)$, we have
$(2-h{)}^2+(-2-k{)}^2=r^2$...(i)
and$(3-h{)}^2,(4k{)}^2-r^2$...(ii)
Also, since the centre lies on the line $x+y=2$, we have
$h+k=2$....(iii)
From [qs. (i) and (ii), we have
$(2-h{)}^2+(-2-k{)}^2=(3-h{)}^2+(4-k{)}^2$
$4+h^2-4h+4+k^2+4k=9+h^2-6h+16+k^2-8k$
$\Rightarrow 4-4h+4+4k=9-6h+16-8k$
$\Rightarrow -4h+6h+4k+8k=9+16-8$
$\to 2h+12k=17$...(iv)
Now, multiplying Eq. (iii) by 2 and subtracting it from Eq. (iv), we get
$(2h+12k)-(2h+2k)=17-4$
$\Rightarrow 10k=13$
$\Rightarrow k=\frac{13}{10}=1\cdot 3$
Substuting the value of $k$ in Eq. (iii), we get
$h+1\cdot 3=2$
$\Rightarrow h=2-1\cdot 3=0\cdot 7$
$\Rightarrow h=0.7$
Now, substituting the value of $h$ and $k$ in Eq. (ii), we get
$(3-0\cdot 7{)}^2+(4-1\cdot 3{)}^2=r^2$
$\Rightarrow 5\cdot 29+7\cdot 29=r^2$
$\text{ or }{r}^2=12\cdot 58$
$\text{ and }h=0.7,k=1.3$
$r^2=12.58$
Hence, the equation of the required circle is
$(x-0.7{)}^2+(y-1.3{)}^2=12.58$