Equation of the circle which touches the lines x = 0, ;y = 0 and … - Vidyadip

MCQ

Equation of the circle which touches the lines $x = 0,\;y = 0$ and $3x + 4y = 4$ is

A
${x^2} - 4x + {y^2} + 4y + 4 = 0$
✓
${x^2} - 4x + {y^2} - 4y + 4 = 0$
C
${x^2} + 4x + {y^2} + 4y + 4 = 0$
D
${x^2} + 4x + {y^2} - 4y + 4 = 0$

✓

Answer

Correct option: B.

${x^2} - 4x + {y^2} - 4y + 4 = 0$

b
(b) Let centre of circle be $(h, k)$.

Since it touches both axes, therefore $h = k = a.$

Hence equation can be ${(x - a)^2} + {(y - a)^2} = {a^2}$

But it also touches the line $3x + 4y = 4$.

Therefore, $\frac{{3a + 4a - 4}}{5} = a \Rightarrow a = 2$

Hence the required equation of circle is

${x^2} + {y^2} - 4x - 4y + 4 = 0$.

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