if a diameter of the circle 2x^2 +2y^2+4x+ky-27=0 passes through the point5,5 and the slope of the diameter is 0.5 find the value of k 回答問題
Question
if a diameter of the circle 2x^2 +2y^2+4x+ky-27=0 passes through the point5,5 and the slope of the diameter is 0.5 find the value of k 回答問題
Solution
The equation of the circle is given by 2x^2 + 2y^2 + 4x + ky - 27 = 0.
First, we can rewrite this equation in the standard form of a circle equation, which is (x - h)^2 + (y - k)^2 = r^2.
Dividing the equation by 2, we get x^2 + y^2 + 2x + (k/2)y - 13.5 = 0.
Completing the square, we get (x + 1)^2 + (y + k/4)^2 = 14.25.
So, the center of the circle is at (-1, -k/4).
The diameter of the circle passes through the center of the circle and the point (5,5).
Since the slope of the diameter is 0.5, we can use the slope formula (y2 - y1) / (x2 - x1) = m to find the value of k.
Substituting the given values, we get (5 - (-k/4)) / (5 - (-1)) = 0.5.
Solving this equation for k, we get k = 20.
So, the value of k is 20.
Similar Questions
In figure, O is the centre of the circle and LN is a diameter. If PQ is a tangent to the circle at K and KLN 30º+ =, find .PKL+
n the given figure PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.
a straight line l intersects the circle 4x^2 +4y^2+18x-12y-2=0 at 2 points p and q.the slope of l is 3/4.if l passes through3,7,find the coordinates of p and q and the equation of the diameter which is perpendicular to l.do with full calculations
straight line 4x+3y+5=0 touches the circle x^2+y^2+4x+ky+11=0 where k is a constant.find the values of k
the equation of a circle is x^2+y^2+kx-8y-(2k-3)=0 and its radius is k+1,wherek>0.find the value of k and hence the centre of the circle.show that the line segment joining P(-6,7)and q(2,1)is a diameter of the circle
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.