Question

The equation of circle passing through the points $(4,1),(6,5)$ and having centre on the line $4x+y=16$ is

• A. $x^2+y^2-2x-24y+15=0$
• B. $x^2+y^2-2x-24y-15=0$
• C. $x^2+y^2-6x-8y+25=0$
• D. $x^2+y^2-6x-8y+15=0$

Answer 1

The correct option is D $x^2+y^2-6x-8y+15=0$

Let the equation of required circle be
$x^2+y^2+2gx+2fy+c=0$

Above circle is passing through the points $(4,1)$, $(6,5)$, we get
$8g+2f+c=-17\cdots \left(1\right)12g+10f+c=-61\cdots \left(2\right)$

Subtracting the above equations gives
$4g+8f=-44\Rightarrow g+2f=-11\cdots \left(3\right)$

Also $(-g,-f)$ lies on $4x+y=16$, we get
$\Rightarrow -4g-f=16\cdots \left(4\right)$
Using equation $\left(3\right)$ and $\left(4\right)$, we get
$\Rightarrow -7g=21\Rightarrow g=-3\Rightarrow f=-4$

Putting value of $g$ and $f$ in $\left(1\right)$, we get
$-24-8+c=-17\Rightarrow c=15$

Hence, the required equation of circle is
$x^2+y^2-6x-8y+15=0$



Alternate solution:
Let the centre be $(h,k)$
We know that
$4h+k=16\Rightarrow k=16-4h\cdots \left(1\right)$
Now, distance between centre and any point on the circle is equal,
$(h-4{)}^2+(k-1{)}^2=(h-6{)}^2+(k-5{)}^2\Rightarrow h^2-8h+16+k^2-2k+1=h^2-12h+36+k^2-10k+25\Rightarrow 4h-44+8k=0\Rightarrow h+2k-11=0$
Using equation $\left(1\right)$, we get
$\Rightarrow h+2(16-4h)-11=0\Rightarrow h+32-8h-11=0\Rightarrow h=3\Rightarrow k=4$

Therefore, the required equation of circle is
$(x-3{)}^2+(y-4{)}^2=(3-4{)}^2+(4-1{)}^2\Rightarrow x^2+y^2-6x-8y+15=0$