What will be the equation of the circle which touches the line x + 2y + 5 = 0 and passes through the point of intersection of the circle x2 + y2 = 1 and x2 + y2 + 2x + 4y + 1 = 0?a)x2 + y2 + 2x + y = 0b)x2 + y2 + x + 2y = 1c)x2 + y2 + x + 2y = 0d)x2 + y2 + 2x + 2y = 1Correct answer is option 'C'. Can you explain this answer?

Question

Answer

To find the equation of the circle, we need to determine its center and radius.

Step 1: Find the intersection point of the two given circles
The two given circles are:
1. x^2 + y^2 = 1
2. x^2 + y^2 - 2x - 4y + 1 = 0

By solving these two equations simultaneously, we can find the intersection points.

Subtracting equation 1 from equation 2, we get:
-2x - 4y + 1 = 0

Rearranging this equation, we get:
2x + 4y = 1

Dividing both sides by 2, we have:
x + 2y = 1/2

Now we have a system of equations:
x + 2y = 1/2
x - 2y = 0

By solving these equations simultaneously, we find the intersection point:
Adding the two equations, we get:
2x = 1/2
x = 1/4

Substituting x = 1/4 into either equation, we get:
1/4 + 2y = 1/2
2y = 1/2 - 1/4
2y = 1/4
y = 1/8

So the intersection point is (1/4, 1/8).

Step 2: Find the equation of the line tangent to the given line
The given line is:
x + 2y + 5 = 0

To find the equation of the line tangent to this line, we need to find the negative reciprocal of the slope of the given line. The given line can be rewritten as:
2y = -x - 5
y = -(1/2)x - 5/2

The slope of this line is -1/2, so the slope of the tangent line is 2.

The equation of the tangent line passing through the point (1/4, 1/8) is:
y - 1/8 = 2(x - 1/4)

Simplifying this equation, we get:
y - 1/8 = 2x - 1/2
y = 2x + 1/8 - 1/2
y = 2x - 3/8

Step 3: Find the equation of the circle
The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2

Since the circle touches the line x + 2y + 5 = 0, the distance between the center of the circle and the line must be equal to the radius of the circle.

The distance from a point (x, y) to a line Ax + By + C = 0 is given by:
d = |Ax + By + C| / sqrt(A^2 + B^2)

For the line x + 2y + 5 = 0, A = 1, B = 2, and C = 5. Substituting the coordinates of the center of the circle (h, k) into the distance formula, we have:
|1h + 2k + 5| / sqrt(1^2 +

Answer

The equation of any circle through the points of intersection of the given circle is,
x² + y² + 2x + 4y + 1 + k(x² + y² – 1) = 0
x² + y² + 2x(1/(k + 1)) + 2*2y/(k + 1) + (1 – k)/(1 + k) = 0
Clearly, the co-ordinates of the center of the circle (1) are, (-1/(1 + k), -2/(1 + k)) and its radius,
= √[(1/(1 + k))² + (2/(1 + k))² – ((1 – k)]/(1 + k))
= √(4 + k²)/(1 + k)
Clearly, the line x + 2y + 5 = 0 is tangent to the circle (1), hence, the perpendicular distance of the line from the center of the circle = radius of the circle
± (-1/(1 + k))– 2(2/(1 + k)) + 5/ √(1² + 2²) = √(4 + k²)/(1 + k)
Or ±(5k/√5) = √(4 + k²)
Or 5k² = 4 + k²
Or 4k² = 4
Or k = 1 [as, k ≠ -1]
Putting k = 1 in (1), equation of the given circle is,
x² + y² + x + 2y = 0

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