Question

# The equation of the circle with centre at , which passes through the point is

## The correct answer is:

### Related Questions to study

### The foot of the perpendicular from on the line is

### The foot of the perpendicular from on the line is

### The foot of the perpendicular from the pole on the line is

### The foot of the perpendicular from the pole on the line is

### The equation of the line parallel to and passing through is

### The equation of the line parallel to and passing through is

### The line passing through the points , (3,0) is

So here we used the concept of the equation of the line passing through two points. Here we also used the trigonometric terms to find the answer using the formulas. So the final solution is .

### The line passing through the points , (3,0) is

So here we used the concept of the equation of the line passing through two points. Here we also used the trigonometric terms to find the answer using the formulas. So the final solution is .

### Statement-I : If then A=

Statement-II : If then

Which of the above statements is true

### Statement-I : If then A=

Statement-II : If then

Which of the above statements is true

### If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of $(−α,a)$ and the other root will be in the interval $(b,α)$.

### If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, one of the roots will be in the interval of $(−α,a)$ and the other root will be in the interval $(b,α)$.

### If be that roots where , such that and then the number of integral solutions of λ is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between

### If be that roots where , such that and then the number of integral solutions of λ is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the number of integral solutions of λ is in between

### If α,β then the equation with roots will be

Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, will be the equation for the roots .

### If α,β then the equation with roots will be

Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, will be the equation for the roots .

### The equation of the directrix of the conic is

### The equation of the directrix of the conic is

### The conic with length of latus rectum 6 and eccentricity is

### The conic with length of latus rectum 6 and eccentricity is

### For the circle centre and radius are

### For the circle centre and radius are

### The polar equation of the circle of radius 5 and touching the initial line at the pole is

### The polar equation of the circle of radius 5 and touching the initial line at the pole is

### The circle with centre at and radius 2 is

### The circle with centre at and radius 2 is

### Statement-I : If

Statement-II :If

Which of the above statements is true

### Statement-I : If

Statement-II :If

Which of the above statements is true