Question

Equation of the circle which touches the line $x = 0$, $y = 0$ and $3x + 4y = 4$ is
A ${x^2} - 4x + {y^2} + 4y + 4 = 0$
B ${x^2} - 4x + {y^2} + 4y - 4 = 0$
C ${x^2} + 4x + {y^2} + 4y + 4 = 0$
D ${x^2} - 4x + {y^2} - 4y + 4 = 0$

Answer 1

Hint: We will come to know that $k = h = a$. Then with the help of a given line, we will find the center and radius of the circle. After all, the calculation will put the center and radius in the standard equation of a circle to get the required equation of a circle.

Complete step by step Solution:
Let $(h,k)$ be the centre
Let $a$ be the radius of the circle
Since it touches both the axes
$ \Rightarrow k = h = a$
$a = \frac{{3h + 4k - 4}}{5} = k = h$
Cross multiplying
$3h + 4k - 4 = 5k$
Shifting $5k$ to the other side
$3h + 4k - 5k - 4 = 0$
After simplifying the above equation, we get
$3h - k = 4$
Shifting $k$ to another side to get a formula for $k$ in terms of $h$
$k = 3h - 4$ (1)
$3h + 4k - 4 = 5h$
Shifting $5h$ to the other side
$5h - 3h - 4k + 4 = 0$
After simplifying the above equation, we get
$2h - 4k + 4 = 0$
Dividing by 2
$h - 2k + 2 = 0$
Putting the value of k
$h - 2(3h - 4) + 2 = 0$
After solving the above equation
$h - 6h + 8 + 2 = 0$
After Simplifying the above equation
$ - 5h + 10 = 0$
$5h = 10$
Dividing by 2
$h = 2$
Putting in the equation (1)
$k = 3 \times 2 - 4$
$k = 6 - 4$
$k = 2$
$k = h = a = 2$
We know the standard equation of Circle
${(x - h)^2} + {(y - k)^2} = {r^2}$
Putting the values $k = h = a = 2$
${(x - 2)^2} + {(y - 2)^2} = {2^2}$
After solving the square in the above equation
${x^2} + 4 - 4x + {y^2} + 4 - 4y = 4$
Shifting 4 to the other side
${x^2} + 4 - 4x + {y^2} + 4 - 4y - 4 = 0$
After canceling like terms, we get
${x^2} - 4x + {y^2} - 4y - 4 = 0$
Hence, the equation of circle is ${x^2} - 4x + {y^2} - 4y - 4 = 0$

Therefore, the correct option is D.

Note: Students should read the question carefully to make the concept clear. They should do calculations correctly to get the required answer.