Question

A circle has its centre in the first quadrant and passes through the points of intersection of the lines ‘x = 2’ and ‘y = 3’. If it makes intercepts of 3 and 4 units on these lines respectively, its equation is
(a) ${{x}^{2}}+{{y}^{2}}-3x-5y+8=0$
(b) ${{x}^{2}}+{{y}^{2}}-4x-6y+13=0$
(c) ${{x}^{2}}+{{y}^{2}}-6x-8y+23=0$
(d)${{x}^{2}}+{{y}^{2}}-8x-9y+30=0$

Answer 1

Hint : First consider the centre of the circle as (a, b) so the general equation of the circle will be ${{x}^{2}}+{{y}^{2}}-2ax-2by+c=0$, then put the values of $x,y$ and put it back to the main equation and substitute the value of ‘c’ in terms of ‘a’ and ‘b’ . Then use the x-intercept and then find the values of ‘b’. Similarly using y-intercept find the value of ‘a’.

Complete step by step solution :
Now, let’s consider a circle with centre be (a, b).
So the equation of circle will be,
${{x}^{2}}+{{y}^{2}}-2ax-2by+c=0...........(1)$
In the question it is given that the circle passes through (2,3).
Therefore, substituting x=2 and y=3 respectively in the equation (1) we get,
${{2}^{2}}+{{3}^{2}}-2a\times 2-2b\times 3+c=0$
$\begin{aligned} & \Rightarrow 13-4a-6b+c=0 \\\ & \therefore c=4a+6b-13 \\\ \end{aligned}$
So now let’s put back value of c in the equation (1) we get,
${{x}^{2}}+{{y}^{2}}-2ax-2by+(4a+6b-13)=0.........(2)$
Now we will find intercept on line x=2, y=3 respectively.
By substituting x=2 in the equation (1) we get,
$4+{{y}^{2}}-4a-2by+(4a+6b-13)=0..........(3)$
In the equation (2) let ${{y}_{1}}$,${{y}_{2}}$be the roots. So, from equation (3),
Sum of the roots will be,
${{y}_{1}}+{{y}_{2}}=2b........(4)$
Product of the roots will be,
${{y}_{1}}{{y}_{2}}=6b-9..........(5)$
Now as we know that length of intercept is (${{y}_{1}}-{{y}_{2}}$), so we will use the identity which is,
$a-b=\sqrt{{{\left( a+b \right)}^{2}}-4ab}$
${{y}_{1}}-{{y}_{2}}$=$\sqrt{{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}-4{{y}_{1}}{{y}_{2}}}$
Now we will substitute the values of $\left( {{y}_{1}}+{{y}_{2}} \right)$and ${{y}_{1}}{{y}_{2}}$to find the value of values of ‘b’ as we are given the value of ${{y}_{1}}-{{y}_{2}}$which is 3.
So,
${{3}^{2}}={{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}-4{{y}_{1}}{{y}_{2}}$
Substituting the values of equation (4) and (5), we get
$\Rightarrow 9={{\left( 2b \right)}^{2}}-4\left( 6b-9 \right)$
$\Rightarrow 4{{b}^{2}}-24b+27=0$
We will solve the quadratic equation. Now by splitting the middle term, we get
$\Rightarrow 4{{b}^{2}}-6b-18b+27=0$
$\Rightarrow 2b(2b-3)-9(2b-3)=0$

& \Rightarrow (2b-9)(2b-3)=0 \\\ & \Rightarrow 2b-9=0,2b-3=0 \\\ & \Rightarrow 2b=9,2b=3 \\\ & \Rightarrow b=\dfrac{9}{2},b=\dfrac{3}{2} \\\ \end{aligned}$$ Hence the values of b is $\dfrac{9}{2}$ or $\dfrac{3}{2}$. By substituting the value of y =3 in the equation (2) we get, ${{x}^{2}}+9-2ax-6b+(4a+6b-13)=0$ ${{x}^{2}}-2ax+4a-4=0.........(6)$ In equation (6) let ${{x}_{1}}$,${{x}_{2}}$ be the roots , Sum of the roots will be, ${{x}_{1}}+{{x}_{2}}=2a......(7)$ Product of the roots will be ${{x}_{1}}{{x}_{2}}=(4a-4).........(8)$ Now as we know the length of intercept is $({{x}_{1}}-{{x}_{2}})$, so we will use the identity which is , $a-b=\sqrt{{{\left( a+b \right)}^{2}}-4ab}$ $$({{x}_{1}}-{{x}_{2}})=\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}-4{{x}_{1}}{{x}_{2}}}$$ Now we will substitute $({{x}_{1}}+{{x}_{2}})$and ${{x}_{1}}{{x}_{2}}$to find the value or values of a as we are given the value of $({{x}_{1}}-{{x}_{2}})=4$ So, $4={{({{x}_{1}}+{{x}_{2}})}^{2}}-4{{x}_{1}}{{x}_{2}}$ $\Rightarrow 4={{(2a)}^{2}}-4(4a-4)$ $\Rightarrow 4={{a}^{2}}-(4a-4)$ $\Rightarrow {{a}^{2}}-4a=0$ $\Rightarrow a(a-4)=0$ Here the value of a is 0 or 4. So now there are 4 such sets of equation can be made by the sets of values of (a,b) which are $$\left( 0,\dfrac{3}{2} \right),\left( 0,\dfrac{9}{2} \right),\left( 4,\dfrac{3}{2} \right)$$ and $\left( 4,\dfrac{9}{2} \right)$. For $\left( 0,\dfrac{3}{2} \right)$ we will substitute in equation (3), we get ${{x}^{2}}+{{y}^{2}}-2ax-2by+(4a+6b-13)=0$ ${{x}^{2}}+{{y}^{2}}-2x(0)-2y\times \dfrac{3}{2}+(4(0)+6\times \dfrac{3}{2}-13)=0$ $\therefore {{x}^{2}}+{{y}^{2}}-3y+(-4)=0$ For $\left( 0,\dfrac{9}{2} \right)$ we will substitute in equation (3), we get ${{x}^{2}}+{{y}^{2}}-2ax-2by+(4a+6b-13)=0$ ${{x}^{2}}+{{y}^{2}}-2x(0)-2y\times \dfrac{9}{2}+(4(0)+6\times \dfrac{9}{2}-13)=0$ $\therefore {{x}^{2}}+{{y}^{2}}-9y+14=0$ For $\left( 4,\dfrac{3}{2} \right)$ we will substitute in equation (3), we get ${{x}^{2}}+{{y}^{2}}-2ax-2by+(4a+6b-13)=0$ ${{x}^{2}}+{{y}^{2}}-2x(4)-2y\times \dfrac{3}{2}+(4(4)+6\times \dfrac{3}{2}-13)=0$ $\therefore {{x}^{2}}+{{y}^{2}}-8x-3y+12=0$ For $\left( 4,\dfrac{9}{2} \right)$ we will substitute in equation (3), we get ${{x}^{2}}+{{y}^{2}}-2ax-2by+(4a+6b-13)=0$ ${{x}^{2}}+{{y}^{2}}-2x(4)-2y\times \dfrac{9}{2}+(4(4)+6\times \dfrac{9}{2}-13)=0$ $\therefore {{x}^{2}}+{{y}^{2}}-8x-9y+30=0$ Only one equation of the pair$\left( 4,\dfrac{9}{2} \right)$ is available in the option. So, the answer is option (d). **Note** : Here the difficulty or confusion a student can face is putting the value of c back into the equation and also selecting the correct values of a and b.