Test: Circle- 2 - JEE MCQ

The equation of the circle  with centre at (h,k) and radius equal to a is (x−h)² +(y−k)² = a²
 When the circle passes through the origin  and centre lies on x− axis 
⇒h = a and k = 0
Then the equation (x−h)²+(y−k)²=a² becomes (x−a)²+y²=a²
If a circle passes through the origin and centre lies on x−axis then the abscissa will be equal to the radius of the circle and the y−co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form 
(x±a)²+y²=a²⇒x²+a² ±2ax+y²=a²
=x² +y² ±2ax=0 is the required equation of the circle.

(x−2)²+(y+1)² = r²

(3,6). lies on it

⇒ 1+49=r²

⇒ r²=50

⇒ x²+4−4x+y²+1+2y=50.

⇒ x²+y²−4x+2y−45=0

Gen equation of a circle is (x−h)²+(y−k)²=r² (3,0),(1,−6),(4,−1) circle passes through these points
∴ These three points should satisfy the equation
of circle.
(3−h)²+k²=r² ___ (I)
(1−h)² +(−6−k)² = r²(1−h)²+(6+k)²=r²  ___ (II)
From (I) & (II)
(3−h)²+k² =(1−h)²+(6+k)²
9+h²−6h+k² =1+h² 2h+36+k²+12k
9−6h=1−2h+36+12k
9−37=4h+12k
−28=4h+12k
h+3k=−7
h=−7−3k ___ (III)
(4−h)²+(−1−k)²=r² ___ (IV)
From (I) & (IV)
(3−h)²+k² = (4−h)² +(1+k)²
9+h² −6h+k²
 =16+h² −8h+1+k²+2k
9−6h=17−8h+2k
9−17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7−3(−4+h)
h=−7+12−3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x−5/4)² +(y+11/4)² =170/16
x²+25/16−10x/4+y² +121/16+22y/4=170/16
Simplifying, we get 2x²+xy²−5x+11y−3=0

For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒ 4 = |c₁−c₂|/(1+3)^1/2
⇒∣c₁−c₂∣ = 8

Let A (a, b) and G (x. y) Now A, G, O are collinear
⇒ x = (2*0 + a)/3
⇒ a = 3x
and similarly b = 3y
Now (a,b) lies on the circle = 9x² + 9y² = 9
=> x² + y² = 1

Equation of circle is given by,

For line y=c to be tangent to the given circle at point (1,1)
It has to pass through (1,1)
⇒ c = 1

Equation of tangent at (x1 , y1) to the circle x² + y² =25 is given by,
xx1 + yy1 = 25. Now comparing it with 3x+4y = 25
We get required point of contact (3,4)

Let equation of tangent with slope =m and point (0,1)
(y−1)=m(x−0)⇒y=mx+1
Intersection point
x²+(mx+1)²−2x+4(mx+1)=0
(1+m²)x²+(−2+6m)x+5=0
For y=mx+1 to be tangent, discriminant =0
(6m−2)²−4×5(1+m²)=0
36m²+4−24m−20m²+20=0
16m²−20m+24=0
⇒ 2m²−3m−2=0
(2m+1)(m−2)=0

Centre and radius of the given circle are C(2,1) and √4+1+20 = 5 respectively.
Now CP=√8²+6²=10. Hence greatest distance of point P from the given circle is =10+r=15

(x-x1)/(x1+g) = (y-y1)/(y1+g)
[x-1/(2)^½]/(1/(2)^½) = [y-1/(2)^½]/(1/(2)^½)
[(2x)^½ -1 *(2)^½]/(2)^½ = [(2x)^½ -1 *(2)^½]/(2)^½
(2x)^½ -1 = (2y)^½ -1
= x-y = 0

Let tangent from origin be y = mx
Using the condition of tangency, we get

The angle between tangents = π/2

Let P(x₁,y₁) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x₁,y1₎ w.r.t circle x2 + y2 = 4 
xx₁ + yy₁ − 4 = 0   ...(1)
Also 3x₁ + 4y1 − 12 = 0
⇒ x₁ + 4/3y₁ − 4 = 0   ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴ Variable chord of contact always passes through (1, 4/3)

Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x−1)²+(y−2)²=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)^½]/2
= 2(3)^1/2 units
Therefore, PC=2(3)^1/2
⇒ PC²=12
⇒ (x−1)²+(y−2)²=12
⇒ x²+y²−2x−4y+5=12
⇒ x²+y²−2x−4y−7=0

Slope of the given chord = −5/2
Slope of the line joining the midpoint on the chord and the centre of the circle = (3+f)/(2+g)
(5/2)[(3+f)/(2+g)] = −1
⇒ 15 + 5f = 4 + 2g
⇒ Locus is 2x − 5y + 11 = 0

The normal line to circle is →y² - 2 y + 4 x -2 xy=0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]^½ 
=(1+1)^½ =(2)^½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0

Radius of the circle = a 
and centre ≡ (b,a)
And circle is touching the x-axis.
The equation of the circle :
x²+y²−2bx−2ay+b² = 0

Internally touch ∴ common tangent is one.

The Required point is the radical centre of the three given circles. The radical axes of the three circles taken in pairs are 3x - 24 = 0,16y + 120 = 0 and - 3x + 16y + 80 = 0. On solving, the required point is (8, -15/2).

touches the another circle

Now, Central first circle will be
And its radius will be 3 units.
Also,centre of second circle
And radius,

As both touches each other
So,