Question 6

Mathematics Circle Hard

The equation of a circle with diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area of 154 sq. units is

(A) x<sup>2</sup> + y<sup>2</sup>&nbsp;+ 6x - 4y - 36 = 0
(B) x<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">2</span>&nbsp;+ y<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">2</span>&nbsp;+ 6x + 4y - 36 = 0
(C) x<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">2</span>&nbsp;+ y<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">2</span>&nbsp;- 6x + 4y + 25 = 0
(D) None of these
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

\[ \text{Given,} \] \[ 2x - 3y + 12 = 0 \quad \text{...(1)} \] \[ x + 4y - 5 = 0 \quad \text{...(2)} \] \[ (1) - 2(2) \text{ gives,} \] \[ y = 2 \] \[ x = -3 \] \[ \text{Center of the circle } (-3,2) \] \[ \text{Area of circle} = 154 \] \[ \pi r^2 = 154 \] \[ r^2 = \frac{154 \times 7}{22} \] \[ r = 7 \] \[ \text{Hence the equation of circle is,} \] \[ (x - (-3))^2 + (y - 2)^2 = 7^2 \] \[ (x + 3)^2 + (y - 2)^2 = 7^2 \] \[ \text{Expanding the equation, we get,} \] \[ x^2 + 6x + y^2 + 13 - 4y = 49 \] \[ x^2 + 6x + y^2 - 4y - 36 = 0 \]