Question

find equation of circle whose center lies in second quadrant , radius 4 units touching x axis and equation of diameter is 2x +y+2 =0

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Solution:

Given,

Let the center of the circle be O(h,k)

The center lies on the second quadrant, hence the center of the circle O(-h,k)

According to the question, radius ( r )= 4units and touches x-axis, hence radius of the circle (r)=k

Hence, the coordinate of centre of the circle is 4.

As we are given diameter of the circle is 2x+y+2=0

And the center is a point lying in the diameter, hence it should satisfy the equation of the diameter,

2x+y+2=0

2×-h+k+2=0

-2h+4+2=0

-2h+6=0

-2h= -6

h= -6/-2

• h=3

As center of the circle lies in the second quadrant, the coordinate of the center is O(-3,4)

(x-h)^2+(y-k)^2=r^2

or, (x+3)^2 +(y-4)^2=4^2

or, x^2+2×3×x+3^2+y^2-2×4×y+4^2=4^2

or, x^2+y^2+6x-8y+9=0

Hence,x^2+y^2+6x-8y+9=0 is the equation of the required circle . HAVE A GREAT DAY AHEAD.