Circles(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
Answer: Option A. -> 1:
A
Let equation of circle be x2+y2 + 2gx + 2fy + c = 0. If (m,1m) lies on this circle, then
m2+1m2+2gm+2f1m+c=0
or m4+2gm3+2fm+cm2+1=0
This is a fourth degree equation in m having m1,m2,m3,m4 as its roots.
Therefore, m1m2m3m4 = product of roots = 11 = 1.
Explanation:-
Answer: Option A. -> 5x2+5y2−10x+30y+49=0:
A
Radius of circle the required circle =∣∣∣2+3–4√5∣∣∣=1√5
Therefore the equation is,
(x−1)2+(y+3)2=15
i.e, x2+y2−2x+6y+1+9=15
i.e, 5x2+5y2−10x+30y+49=0.
Explanation:-
Answer: Option B. -> Collinear:
B
Centre are (0, 0), (-3, 1) and (6, -2) and a line passing through any two points say (0, 0) and (-3, 1) is y=−13x and point (6, -2) lies on it. Hence points are collinear.
Explanation:-
Answer: Option D. -> y2−10x−6y+14=0:
D
Let (x, y) be centre of circle which touch y - axis and given circle.
At any point, the radius of this circle will be equal to 'x' units. Since this circle is touching the given circle which has a radius of two units,
∴ Distance between centres of the two circles = 2 + x
⇒√(x−3)2+(y−3)2=2+x
y2−10x−6y+14=0
Explanation:-
Answer: Option B. -> 10:
B
Inclination of the line ←→AB is 135∘⇒Slope=tan135∘=−1
Equation of ←→AB is y−√8=−1(x+√8)⇒x+y=0
x + y = 0 passes through the centre of the circle x2+y2=25
∴ Length of the chord AB = Diameter of the circle =2×5=10
Explanation:-
Answer: Option D. -> p∈ϕ:
D
Since the point P is interior to the circle, S1 < 0
=(22)+(82)−2.(2)+4(8)−p<0
=96−p<0
=p>96
Also given that the circle doesn't touches any of the axes.
So, g2−c < 0
f2−c < 0
g2−c < 0
=1+p<0
=p<−1
Also,
f2−c < 0
=4+p<0
=p<−4
Since p<−4 and p>96 is not possible, the set for p will be a null set meaning it'll not have any element in it.
Explanation:-
Answer: Option C. -> 1c:
C
We can write the equation of common tangent, with the help of radical axis.
Since two circles are touching each other the common tangent will be the radical axis, and the equation of the radical axis we can write -
S = S'
x2+y2+2ax+c=x2+y2+2by+c=0
So, Equation of Common tangent is ax – by = 0.
Now, we can find the perpendicular distance from the center of the first circle to the tangent which also should be equal to radius of that circle.
Perpendicular distance from (-a,0) to the tangent = a2√a2+b2
Radius of the same circle will be = √a2−c
Therefore a2√a2+b2=√a2−c