
The equation of the image of the circle ${x^2} + {y^2} + 16x - 24y + 183 = 0$ by the line mirror 4x+7y+13=0 is:
A. ${x^2} + {y^2} + 32x - 4y + 235 = 0$
B. ${x^2} + {y^2} + 32x + 4y - 235 = 0$
C. ${x^2} + {y^2} + 32x - 4y - 235 = 0$
D. ${x^2} + {y^2} + 32x + 4y + 235 = 0$
Answer
493.2k+ views
Hint: In this question, to find the image of the circle, we will find the image of the centre of the circle by the given line and radius will be the same. To find the image of a point ,we will write the equation of the line in symmetrical form and then find the image of the centre point of the circle.
Complete step by step answer:
The given equation of circle is ${x^2} + {y^2} + 16x - 24y + 183 = 0$ , we will convert this in standard form to find the centre of the circle.
The given circle equation can be written as:
${x^2} + 16x + 64 + {y^2} - 24y + 144 + - 25 = 0$
$ \Rightarrow {(x + 8)^2} + {(y - 12)^2} = 25 = {5^2}$ (1)
We know that the standard form of circle equation is given as:
$ \Rightarrow {(x - a)^2} + {(y - b)^2} = {r^2}$ (2)
Comparing equation 1 and 2, we have:
Centre: (-8 , 12) , radius = 5cm.
Image of the centre by line 4x+7y+13 =0 is given by:
$\dfrac{{x - ( - 8)}}{4} = \dfrac{{y - 12}}{7} = \dfrac{{(4( - 8) + 7(12) + 13 \times ( - 2))}}{{{4^2} + {7^2}}}$
On simplifying the above expression, we get:
$\dfrac{{x + 8}}{4} = \dfrac{{y - 12}}{7} = \dfrac{{65 \times - 2}}{{65}}$
$ \Rightarrow \dfrac{{x + 8}}{4} = - {\text{ and }}\dfrac{{y - 12}}{7} = - 2$
$ \Rightarrow $ x= -16 and y = -2.
Therefore, the new centre is ( -16,-2)
The radius is same as earlier i.e. r = 5cm
Therefore, the new equation of circle which is formed by taking the image of the given circle by the line 4x+7y+13=0 is given as:
$ \Rightarrow {(x + 16)^2} + {(y + 2)^2} = {5^2} = 25$
${x^2} + {y^2} + 32x + 4y + 235 = 0$
Hence, option D is the correct option.
Note:
In this type of question, you should know how to find the image of a point by a given line. The image of a point can also be found by assuming a point which lies on the given line and line joining the point and its image and then using the concept that point on the given line is mid point of given point and its image also the two lines are perpendicular.
Complete step by step answer:
The given equation of circle is ${x^2} + {y^2} + 16x - 24y + 183 = 0$ , we will convert this in standard form to find the centre of the circle.
The given circle equation can be written as:
${x^2} + 16x + 64 + {y^2} - 24y + 144 + - 25 = 0$
$ \Rightarrow {(x + 8)^2} + {(y - 12)^2} = 25 = {5^2}$ (1)
We know that the standard form of circle equation is given as:
$ \Rightarrow {(x - a)^2} + {(y - b)^2} = {r^2}$ (2)
Comparing equation 1 and 2, we have:
Centre: (-8 , 12) , radius = 5cm.
Image of the centre by line 4x+7y+13 =0 is given by:
$\dfrac{{x - ( - 8)}}{4} = \dfrac{{y - 12}}{7} = \dfrac{{(4( - 8) + 7(12) + 13 \times ( - 2))}}{{{4^2} + {7^2}}}$
On simplifying the above expression, we get:
$\dfrac{{x + 8}}{4} = \dfrac{{y - 12}}{7} = \dfrac{{65 \times - 2}}{{65}}$
$ \Rightarrow \dfrac{{x + 8}}{4} = - {\text{ and }}\dfrac{{y - 12}}{7} = - 2$
$ \Rightarrow $ x= -16 and y = -2.
Therefore, the new centre is ( -16,-2)
The radius is same as earlier i.e. r = 5cm
Therefore, the new equation of circle which is formed by taking the image of the given circle by the line 4x+7y+13=0 is given as:
$ \Rightarrow {(x + 16)^2} + {(y + 2)^2} = {5^2} = 25$
${x^2} + {y^2} + 32x + 4y + 235 = 0$
Hence, option D is the correct option.
Note:
In this type of question, you should know how to find the image of a point by a given line. The image of a point can also be found by assuming a point which lies on the given line and line joining the point and its image and then using the concept that point on the given line is mid point of given point and its image also the two lines are perpendicular.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How do I convert ms to kmh Give an example class 11 physics CBSE

Describe the effects of the Second World War class 11 social science CBSE

Which of the following methods is suitable for preventing class 11 chemistry CBSE
