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Q.
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
Conic Sections
Solution:
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$(3 - h)^2,(4 k)^2-r^2$...(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^2-4 h+4+k^2+4 k =9+h^2-6 h+16+k^2-8 k$
$\Rightarrow 4-4 h+4+4 k =9-6 h+16-8 k$
$\Rightarrow -4 h+6 h+4 k+8 k =9+16-8$
$\rightarrow 2 h+12 k =17$...(iv)
Now, multiplying Eq. (iii) by 2 and subtracting it from Eq. (iv), we get
$(2 h+12 k)-(2 h+2 k) =17-4 $
$\Rightarrow 10 k =13$
$\Rightarrow k =\frac{13}{10}=1 \cdot 3$
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
$(x-0.7)^2+(y-1.3)^2=12.58$