Did you know that two different circles can touch both coordinate axes and pass through the same point? The tangency conditions fix the centre’s distance from each axis to equal the radius. For a circle lying in the first quadrant, its centre must therefore be (r,r), with r > 0. Substitute a specified point into the circle equation and solve a quadratic for r.
For example, find every circle tangent to both coordinate axes and passing through P(1,2). The centre is (r,r), and the point condition gives (1−r)² + (2−r)² = r². This simplifies to (r−1)(r−5) = 0, giving radii 1 and 5. The two valid circles are (x−1)² + (y−1)² = 1 and (x−5)² + (y−5)² = 25. This guide derives the method, verifies both answers and handles other quadrants, axis points and impossible conditions.
What does tangent to both coordinate axes mean?
The coordinate axes are the two perpendicular straight lines x = 0 and y = 0.
A circle tangent to the x-axis touches y = 0 at exactly one point. Its centre’s perpendicular distance to the x-axis equals its radius.
Similarly, tangency to the y-axis requires the centre’s perpendicular distance to x = 0 to equal the same radius.
These are two geometric constraints on the centre. A circle equation cannot be chosen by guessing a picture; the centre-to-line distance must satisfy both constraints exactly. The tangency conditions are what make the circle’s centre especially simple to represent.
Translate the tangency conditions into centre coordinates
Let the centre be O(h,k) and radius r > 0. Distance from O to the vertical y-axis x = 0 is |h|, so tangency there requires |h| = r.
Distance from O to the horizontal x-axis y = 0 is |k|, so tangency there requires |k| = r.
Therefore h = ±r and k = ±r. A centre can occupy any of four quadrants, giving forms (r,r), (−r,r), (−r,−r) or (r,−r).
A point specified in a particular open quadrant helps select the appropriate centre signs. Do not automatically assume centre (r,r) when the point lies below the x-axis or to the left of the y-axis.
Why a first-quadrant point selects the first-quadrant circle
A circle with centre (r,r) and radius r lies in x ≥ 0 and y ≥ 0, touching both axes at the boundaries.
Similarly, a circle centred at (−r,r) lies in the second quadrant together with its boundary axes, and so on for the other sign combinations.
A circle tangent to both axes cannot reach a point strictly inside a different quadrant because its disk does not cross either axis.
Therefore when the supplied point P(p,q) has p > 0 and q > 0, any circle through it that is tangent to both axes must have centre (r,r), with r > 0. The sign selection is geometry, not a convenient guess.
Set up the opening example
Let P(1,2) and suppose the circle touches both coordinate axes. Because P lies strictly in Quadrant I, take centre O(r,r), with radius r > 0.
The circle equation is (x−r)² + (y−r)² = r². Substitute the point x = 1, y = 2.
This gives (1−r)² + (2−r)² = r². Expand the two squared brackets carefully before collecting the terms.
One known point has turned the family of first-quadrant tangent circles into a quadratic equation for the unknown radius. Every positive radius solving that equation gives a candidate circle.
Solve the radius equation
Expand (1−r)² = 1 − 2r + r² and (2−r)² = 4 − 4r + r².
Add them and set the result equal to r²: 5 − 6r + 2r² = r².
Rearrange to r² − 6r + 5 = 0. Factorise as (r−1)(r−5) = 0.
Therefore r = 1 or r = 5. Both values are positive, so both can represent genuine circle radii. Neither should be discarded merely because one is much larger than the other.
Write the smaller circle exactly
For r = 1, the centre is O₁(1,1). Its radius is 1, so its equation is (x−1)² + (y−1)² = 1.
The horizontal x-axis touches this circle at (1,0), directly below the centre. The vertical y-axis touches at (0,1), directly left of the centre.
The supplied point P(1,2) also lies on its circumference: (1−1)² + (2−1)² = 0 + 1 = 1.
Thus the smaller circle satisfies both tangency conditions and the point condition. It is one complete valid answer to the reconstruction problem.
Write the larger circle exactly
For r = 5, the centre is O₂(5,5), with radius 5. Its equation is (x−5)² + (y−5)² = 25.
It touches the x-axis at (5,0) and the y-axis at (0,5). Those points are at distance 5 from the centre along perpendicular radii.
Substitute P(1,2): (1−5)² + (2−5)² = (−4)² + (−3)² = 16 + 9 = 25.
Therefore P lies on the larger circle as well. The 3–4–5 distance verifies why a circle much larger than the supplied point’s coordinates can still pass through that point.
Why both circles are geometrically possible
The smaller circle bends tightly around the point P(1,2), touching the axes close to the origin. Its centre is only one unit from either axis.
The larger circle has centre (5,5) and radius 5. It curves back toward P while touching the axes at more distant contact points.
A sketch may initially suggest only the small circle, but the quadratic radius equation proves that two positive values satisfy the conditions. Both are legitimate, and neither is a duplicate of the other.
This is an important lesson about reconstruction: one point and two tangency conditions may determine a finite set of possible objects rather than exactly one. Report every admissible solution.
Verify the tangent points by perpendicular radii
A circle centred at (r,r) touches the x-axis at (r,0). The radius from (r,r) to (r,0) is vertical and hence perpendicular to the horizontal x-axis.
It touches the y-axis at (0,r). That radius is horizontal and therefore perpendicular to the vertical y-axis.
The distance to either contact point is exactly r. Thus any circle (x−r)² + (y−r)² = r² with r > 0 is tangent to both axes.
This verifies the family used in the algebra. The two answers do not merely have their centres on y = x; they satisfy the defining tangent-radius relationship at both axes.
Derive the general equation for a first-quadrant point
Let P(p,q) satisfy p > 0 and q > 0. A circle through P tangent to both coordinate axes has centre (r,r) and radius r > 0.
The point condition is (p−r)² + (q−r)² = r².
Expand: p² − 2pr + r² + q² − 2qr + r² = r².
Collect terms to obtain r² − 2(p+q)r + p² + q² = 0. This is the general quadratic that determines the possible radii from a first-quadrant point.
Solve the general radius equation
Apply the quadratic formula to r² − 2(p+q)r + p² + q² = 0.
The discriminant is [−2(p+q)]² − 4(p²+q²) = 4[(p+q)² − (p²+q²)] = 8pq.
Therefore r = [2(p+q) ± √(8pq)]/2 = p + q ± √(2pq).
For a first-quadrant point with p,q > 0, the two values are distinct. Both are positive because p+q ≥ 2√(pq) > √(2pq). Hence exactly two first-quadrant tangent circles pass through every point strictly inside Quadrant I.
Recover the opening radii using the general formula
For P(1,2), let p = 1 and q = 2. The general formula gives r = 1 + 2 ± √(2·1·2).
Thus r = 3 ± √4 = 3 ± 2, giving r = 1 or r = 5.
These are exactly the two radius values obtained by direct factorisation. The agreement is a strong check on the expansion and formula.
The sum of the two radii is 2(p+q) = 6, and their product is p² + q² = 5. Indeed 1 + 5 = 6 and 1·5 = 5, as expected from the quadratic coefficient relationships.
A second example with non-integer radii
Suppose the circle passes through P(3,4) in Quadrant I and is tangent to both axes.
The general formula gives r = 3 + 4 ± √(2·3·4) = 7 ± √24 = 7 ± 2√6.
Both numbers are positive because 7 > 2√6. They give two different valid first-quadrant circles, centred at (r,r) with radius r.
The exact equations are (x−r₁)² + (y−r₁)² = r₁² and (x−r₂)² + (y−r₂)² = r₂², where r₁ = 7−2√6 and r₂ = 7+2√6. Exact surds avoid unnecessary rounding of centres and radii.
Why the radius solutions are always positive in the open quadrant
For p,q > 0, the arithmetic–geometric mean inequality gives p+q ≥ 2√(pq).
Since 2√(pq) > √(2pq) for positive pq, we have p+q − √(2pq) > 0.
Therefore even the smaller general radius is strictly positive. The larger one is obviously positive.
This proves both quadratic roots represent admissible radii, not merely real algebraic numbers. It is stronger than checking positivity in one numerical example and confirms that a generic first-quadrant point yields two positive-radius solutions.
The two radii have a simple sum and product
The radius equation is r² − 2(p+q)r + p² + q² = 0. The two roots therefore satisfy r₁ + r₂ = 2(p+q).
Their product is r₁r₂ = p² + q², the squared distance from P(p,q) to the origin.
For P(1,2), the squared origin distance is 1² + 2² = 5, agreeing with 1·5 = 5.
This offers an elegant connection between circle reconstruction and quadratic root identities. One geometric observation—the fixed distance from P to the origin—reappears as the product of the two possible radii.
What the larger and smaller circles have in common
Both circles touch the same two perpendicular axes and pass through the specified point, but their centres and tangent contacts differ.
The smaller circle uses r₁ = p+q−√(2pq), while the larger uses r₂ = p+q+√(2pq). Their centres lie on the diagonal line y = x.
The two radii are symmetric around p+q, because their arithmetic mean is p+q. Their difference is 2√(2pq).
A problem that asks for only one circle must supply an extra restriction, such as radius less than a specified value, centre location in a given region or the particular tangent contact coordinate. Without that, reporting both circles is correct.
A specified radius bound can select one circle
Return to P(1,2), which allows radii 1 and 5. If the question adds r < 3, only the smaller circle is admissible.
If it instead adds r > 3, only the larger circle is admissible. If it says r = 3, no circle meets all the conditions because 3 is not a root of r²−6r+5.
This is how an additional piece of geometric data can resolve an otherwise two-solution reconstruction problem.
The correct procedure is to find all candidates first and then filter them using the stated additional constraint. Do not build an unstated size preference into the initial algebra.
A point in another quadrant changes the centre signs
Suppose P lies strictly in Quadrant II, with p < 0 and q > 0. Any circle tangent to both axes through it must lie in that quadrant, with centre (−r,r).
For a point in Quadrant IV, use centre (r,−r). For Quadrant III, use (−r,−r). The radius is always positive.
Within the appropriate quadrant, the same formula works using the positive coordinate magnitudes u = |p| and v = |q|: r = u+v ± √(2uv).
This means the two-solution structure is preserved in every open quadrant. What changes is the signs of the centre coordinates, not the positive radius calculation.
A fourth-quadrant worked example
Let P(1,−2). Because it lies in Quadrant IV, a circle through it tangent to both axes must have centre (r,−r), with r > 0.
The point condition is (1−r)² + (−2+r)² = r². Squaring removes the sign of the second bracket, producing the same equation r²−6r+5 = 0.
Thus r = 1 or 5. The circle equations are (x−1)² + (y+1)² = 1 and (x−5)² + (y+5)² = 25.
Both lie in the fourth quadrant, touch the appropriate positive x-axis and negative y-axis directions, and pass through (1,−2). A reflection across the x-axis connects them to the original example.
A point on an axis is a boundary case
Suppose the given point is P(3,0) on the positive x-axis. If a circle is tangent to the x-axis and passes through P, then P must be that point of tangency.
Its centre must have x-coordinate 3 and vertical coordinate 3 or −3 to remain three units from both axes. Therefore there are two circles: (x−3)² + (y−3)² = 9 and (x−3)² + (y+3)² = 9.
They lie on opposite sides of the x-axis. The formula for a first-quadrant interior point, if extended algebraically with q = 0, gives a repeated radius r = 3, but there are two possible centre-sign choices at this boundary.
This is why the special case of a point lying on an axis must be treated separately from a point strictly inside one quadrant.
The origin cannot lie on a positive-radius circle tangent to both axes
Consider P(0,0), the intersection of the axes. A circle tangent to both axes with radius r > 0 has centre (±r,±r).
The distance from any such centre to the origin is √(r²+r²) = r√2, strictly greater than r.
Therefore the origin lies outside the circumference, not on it. No positive-radius circle can be tangent to both coordinate axes and pass through their intersection.
The algebraic radius equation at p = q = 0 gives r² = 0, whose only root r = 0 is inadmissible for a genuine circle. This is a useful example of why a quadratic root must be checked against the geometric requirement r > 0.
What if tangency is only to one axis?
If a circle need only touch the x-axis, its centre can have any horizontal coordinate h and vertical coordinate ±r. The y-axis imposes no additional distance condition.
Passing through one specified point gives one equation in h and r, so in general infinitely many circles may remain possible.
Tangency to both axes is substantially stronger: it forces both centre-coordinate magnitudes to equal r. That reduces the unknowns to one positive radius within a selected quadrant.
This distinction explains why the two-axis problem yields only two candidates for an interior point whereas a one-axis problem usually needs another condition for uniqueness.
A circle may be tangent without its centre being on either axis
For centre (r,r), the radius to the x-axis contact point (r,0) is vertical, and the radius to the y-axis contact point (0,r) is horizontal.
The centre itself lies on the diagonal y = x, not on either coordinate axis for r > 0.
A student who sets the centre to (r,0) because the circle touches y = 0 has confused the contact point with the centre. Such a circle would be centred on the x-axis and would generally intersect it at two points, not be tangent to it.
Mark the contact points and centre separately on a sketch. Their geometry explains the coordinate constraints much better than copying a centre formula without understanding it.
Why the contact points do not coincide with the supplied point
In the opening example P(1,2), neither P is on an axis, so it cannot be either axis contact point.
For the smaller circle, the tangent contacts are (1,0) and (0,1). For the larger circle, they are (5,0) and (0,5).
The given point P lies elsewhere on both circumferences. It supplies the remaining radius equation, not a contact coordinate.
If a problem instead gives an axis point as the exact contact point, that extra geometric meaning can determine the horizontal or vertical centre coordinate directly. Read whether a point is stated as “on the circle” or “point of tangency”.
Check one circle using distance to both axes
Take the larger circle with centre (5,5) and radius 5.
The distance from (5,5) to the x-axis y = 0 is |5| = 5, equal to the radius. The distance to the y-axis x = 0 is also |5| = 5.
Therefore each axis is tangent. The circle does not cross into a region with negative x or negative y because its smallest possible x and y coordinates are both zero.
This is an independent verification of the geometric conditions and helps establish that the solution is truly a first-quadrant tangent circle, not just a circle whose equation happens to contain two matching coordinate shifts.
A quadratic equation may have two roots but only one valid radius in another problem
In circle reconstruction generally, a quadratic equation for the radius may produce a negative value, zero or a value violating a stated geometry constraint.
A genuine radius must be positive. A candidate with r = 0 describes a degenerate point, while r < 0 has no direct meaning as a circle radius.
For a point strictly inside a quadrant and tangency to both axes, we proved both solutions are positive, but that does not justify skipping the positivity check in every different circle problem.
The correct habit is to solve the algebraic equation first and then verify every candidate against the original distance, sign and domain conditions.
A compact six-step exam routine
First identify the point’s quadrant. For an interior first-quadrant point, write centre (r,r) and radius r > 0.
Second substitute the given point into (x−r)²+(y−r)²=r². Third expand and solve the resulting quadratic in r.
Fourth reject invalid candidates according to positivity, quadrant and any additional stated radius restrictions. Fifth write a separate circle equation for every admissible radius.
Sixth verify the given point and confirm that the distance from the centre to each coordinate axis equals r. Boundary-axis and origin cases require separate sign or degenerate-case reasoning. This completes the reconstruction without relying on visual guesses.
Practice questions with checked answers
Question one: find circles tangent to both axes through P(1,2). Answers: (x−1)²+(y−1)²=1 and (x−5)²+(y−5)²=25.
Question two: for P(3,4) in Quadrant I, find the possible radii. Answer: 7−2√6 and 7+2√6.
Question three: for P(p,q) with p,q > 0, give the general radii. Answer: p+q ± √(2pq).
Question four: find circles tangent to both axes through P(1,−2). Answers: (x−1)²+(y+1)²=1 and (x−5)²+(y+5)²=25.
Question five: find circles tangent to both axes through P(3,0). Answers: (x−3)²+(y−3)²=9 and (x−3)²+(y+3)²=9.
Question six: is there a positive-radius circle tangent to both axes through (0,0)? No.
For each, identify the permitted centre quadrants before solving for radius.
Parent guidance, syllabus scope and related reading
This topic joins circle equations, centre-to-line distance, quadrant geometry, quadratic factorisation and information sufficiency. Multiple-solution and axis-boundary variations can be taught as enrichment according to a student’s school sequence.
For Singapore, consult the relevant SEAB syllabus listings for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2, as relevant.
Related guides include How to Find a Circle Tangent to Two Parallel Lines, How to Find a Circle Through Two Points with Its Centre on a Given Line, How to Find a Circle’s Tangency Point Using a Perpendicular Foot and the Additional Mathematics archive.
Parents may enquire about suitable eduKate Singapore small-group Mathematics lessons near Sixth Avenue MRT. The lasting skill is to translate geometric tangency into exact distance conditions and recognise when those conditions legitimately produce two different circles. Properly taught kids shine a bright light into the future.